ahlfors l.v. - lectures on quasi con formal mappings (van nostrand, 1966)

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Lars V. Ahlfors-LECTURES ON QUASICONFORMAL MAPPINGS

Additional titles will be listed and announced as issued.

Lectures on

QUASICONFORMAL MAPPINGS

byLARS v: AHLFORS

Harvard University

Manuscript prepared with the assistance of

CLIFFORD J. EARLE Jr.Cornell University

D. VAN NOSTRAND COMPANY, INC.

PRINCETON, NEW JERSEY

TORONTO NEW YORK LONDON

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COPYRIGHT © 1966, BYD. VAN NOSTRAND COMPANY, INC.

Published simultaneously in Canada byD. VAN NOSTRAND COMPANY (Canada), LTD.

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PRINTED IN THE UNITED STATES OF AMERICA

ACKNOWLEDGMENTS

The manuscript was prepared by Dr. Clifford Earle from

rough long hand notes of the author. He has contributed many

essential corrections, checked computations and supplied many

of the bridges that connect one fragment of thought with the next.

Without his devoted help the manuscript would never have attain

ed readable form.

In keeping with the informal character of this little vol

ume there is no index and the references are very spotty, to say

the least. The experts will know that the history of the subject

is one of slow evolution in which the authorship of ideas cannot

always be pinpointed.

The typing was excellently done by Mrs. Caroline W.

Browne in Princeton, arid financed by Air Force Grant AFOSR

393-63.

CONTENTS

CHAPTER I: DIFFERENTIABLE QUASICONFORMAL

MAPPINGS

Introduction . . . . . . . . . . . . . . . . . . . . . . . .. 1

A. The Problem and Definition of Grotzsch . . .. 2B. Solution of Grotzsch's Problem. . . . . . . . .. 6

C. Composed Mappings. . . . . . . . . . . . . . . .. 8

D. Extremal Length " 10

E. A Symmetry Principle . . . . . . . . . . . . . . .. 16

F. Dirichlet Integrals. . . . . . . . . . . . . . . . .. 17

CHAPTER II: THE GENERAL DEFINITION

A. The Geometric Approach . . . . . . . . . . . . .. 21

B. A 1-q-c. map is Conformal. . . . . . . . . . . .. 23

CHAPTER III: EXTREMAL GEOMETRIC PROPERTIES

A. Three Extremal Problems. . . . . . . . . . . . .. 35

B. Elliptic and Modular Functions. . . . . . . . .. 40

C. Mori's Theorem. . . . . . . . . . . . . . . . . . .. 47

D. Quadruplets....................... 53

CHAPTER IV: BOUNDARY CORRESPONDENCE

A. The M-condition. . . . . . . . . . . . . . . . . . .. 63

B. The Sufficiency of the M-condition. . . . . . .. 69

C. Quasi-isometry . . . . . . . . . . . . . . . . . . .. 73

D. Quasiconformal Reflection. . . . . . . . . . . .. 74

E. The Reverse Inequality. . . . . . . . . . . . . .. 81

Quasiconformal Mappings

CHAPTER V: THE MAPPING THEOREM

A. Two Integral Operators. . . . . . . . . . . . . .. 85

B. Solution of the Mapping Problem. . . . . . . .. 90

C. Dependence on Parameters . . . . . . . . . . .. 100

D. The Calder6n-Zygmund Inequality. 106

CHAPTER VI: TEICHMULLER SPACES

A. Preliminaries . . . . . . . . . . . . . . . . . . . .. 117

B. Beltrami Differentials 120

C. !!J. is Open 128

D. The Infinitesimal Approach . . . . . . . . . . .. 137

CHAPTER I

DIFFERENTIABLE QUASICONFORMAL MAPPINGS

Introduction

There are several reasons why quasiconformal mappings

have recently come to playa very active part in the theory of

analytic functions of a single complex variable.

1. The most superficial reason is that q.c. mappings are

a natural generalization of conformal mappings. If this were their

only claim they would soon have been forgotten.

2. It was noticed at an early stage that many theorems

on conformal mappings use only the quasiconformality. It is

therefore of some interest to determine when conformality is es

sential and when it is not.

3. Q.c. mappings are less rigid than conformal mappings,

and are therefore much easier to use as a tool. This was typi

cal of the utilitarian phase of the theory. For instance, it was

used to prove theorems about the conformal type of siinply con

nected Riemann surfaces (now mostly forgotten).1

2 Quasiconformal Mappings

4. Q.c. mappings play an important role in the study of

certain elliptic partial differential equations.

5. Extremal problems in q.c. mappings lead to analytic

functions connected with regions or Riemann surfaces. This

was a deep and unexpected discovery due to Teichmiiller.

6. The problem of moduli was solved with the help of

q.c. mappings. They also throw light on Fuchsian and Kleinian

groups.

7. Conformal mappings degenerate when generalized to

several variables, but q.c. mappings do not. This theory is

still in its infancy.

A. The Problem and Definition of Grotzsch

The notion of a quasiconformal mapping, but not the

name, was introduced by H. Grotzsch in 1928. If Q is a

square and R is a rectangle, not a square, there is no confor

mal mapping of Q on R which maps vertices on vertices.

Instead, Grotzsch asks for the most nearly conformal mapping

of this kind. This calls for a measure of approximate confor

mality, and in supplying such a measure Grotzsch took the

first step toward the creation of a theory of q.c. mappings.

All the work of Grotzsch was late to gain recognition,

and this particular idea was regarded as a curiosity and allowed

to remain dormant for several years. It reappears in 1935 in the

. work of Lavrentiev, but from the point of view of partial differ

ential equations. In 1936 I included a reference to the q.c. case

Differentiable Quasiconformal Mappings

in my theory of covering surfaces. From then on the notion

became generally known, and in 1937 Teichmiiller began to

prove important theorems by use of q.c. mappings, and later

theorems about q.c. mappings.

3

We return to the definition of Gr6tzsch. Let w = f(z)

(z = x + iy, w = u + iv) be a C1 homeomorphism from one re

gion to another. At a point Zo it induces a linear mapping of

the differentials

(1)du

dv

u dx + u dyx y

v dx + v dyx y

which we can also write in the complex form

(2)

with

dw = f dz + f- dz?: Z

(3) f-=';Ii(f+if)z x y

Geometrically, (1) represents an affine transformation

from the (dx, dy) to the (du, dv) plane. It maps circles a

bout the origin into similar ellipses. We wish to compute the

ratio between the axes as well as their direction.

In classical notation one writes

(4)

with

G

The eigenvalues are determined from


(5)

and are

\E- A P I

P G- A o

(6) \ \ _ E+G+[(E_G)2+4p2]'h1\1' 1\2 - 2

(7)

The ratio a: b of the axes is

(Al)lh = E+ G+[(E_G)2+4p 2]'h

A2 2(EG _ p 2) 'h

The complex notation is much more convenient. Let us

first note that

(8)

This gives

f-z

lh (u + v ) + j~ (v - u )x y x y

yiu - v ) + j~ (v + u )x y x y

(9) If 12 - \ f_1 2 = u v - u v = ]z z xy yx

which is the Jacobian. The Jacobian is positive for sense pre

serving and negative for sense reversing mappings. For the

moment we shall consider only the sense preserving case.

Then 1f'Z I < 1fz 1 .

It now follows immediately from (2) that

where both limits can be attained. We conclude that the ratio

of the major to the minor axis is

(11)If 1+ If-ID - z z > 1

f - 1f zl - If'Z 1

Differential Quasiconformal Mappings

This is called the dilatation at the poiJ1t z. It is often more

convenient to consider

5

(12)

related to D l by

(13)

The mapping is conformal at z if and only if Dl

The maximum is attained when the ratio

1, dl = O.

f- clZz

f zdz

is positive, the minimum when it is negative. We introduce now

the complex dilatation

(14)

with Illll

(15)

fIII = f z

zdl . The maximum corresponds to the direction

arg dz = a = 'h arg fl

the minimum to the direction a ±"/2. In the dw-plane the di

rection of the major axis is

(16)

where we have set

arg dw 'h arg v

(17)

The quantity v l may be called the second complex dilatation.

6

figure:


We will illustrate by the following self-explanatory

Observe that (3 - a = arg f z'

Definition 10 The mapping f is said to be quasiconfor

mal if D f is bounded. It is K -quasiconformal if D f S K.

The condition D f SKis equivalent to df S k =

(K -l)/(K + 1). A l-quasiconformal mapping is conformal.

Let it be said at once that the restriction to C1-mappings

is most unnatural. One of our immediate aims is to get rid of

this restriction. For the moment, however, we prefer to push

this difficulty aside.

B. Solution of Grotzsch's Problem

We pass to Grotzsch's problem and give it a precise

meaning by saying that f is most nearly conformal if sup D f

is as small as possible.

Let R, R' be two rectangles with sides a, b and a ~

b '. We may assume that a: b sa': b ' (otherwise, interchange

a and b). The mapping f is supposed to take a-sides into

a-sides and b-sides into b-sides.

Differentiable Quasiconiormal Mappings 7

60 6' Ia a'

The computation goesa a

,S f Idi(x+ iY)1 f (Iizl + liz I) dxa <

0 0

a b

a'b < f f (Iizl + li~) dx dy

o 0

lizl + lizllizl -lizl

a b

dx dy f fo 0

a b

a'b ' f f D f dx dy

o 0

or

(1) f f Dfdx dy

R

and in particular

~. ~ < sup Dfb' . b -

The minimum is attained for the affine mapping which

is given by

Hz)

8 Quasiconiormal Mappings

THEOREM 1. The affine mapping has the least maximal

and the least average dilatation.

The ratios m = alband m' = a'Ib' are called the

modhles of R and oR' (taken with an orientation). We have

proved that there exists a K-q.c. mapping of R on R' if and

only if

(2)

C. Composed Mappings

We shall determine the complex derivatives and complex

dilatations of a composed mapping g 0 i. There is the usual

trouble with the notation which is most easily resolved by in

troducing an intermediate variable ,= f(z).

The usual rules are applicable and we find

(1)(g 0 f) z = (g, 0 f) iz + (g, 0 f) 7z

(~ 0 f)- = (g 0 f) i- + (g- 0 f) 7-5 Z , Z , Z

When solved they give

(2)

where

1 -- [(g 0 f) i-1 z z

g- 0 i 1 [(g 0 f)- i, 1 z z

2 21 = Iizl - Iii I .

- (g 0 f) -7 ]z z

- (g 0 f) i-]z z

For g = i-I the formulas become

(3) (i-I) 0 i = 7-11 ,, z

Differ€f1tiable Quasicon£ormal Mappings 9

One derives, for instance,

(4) £-1/l -1 = -1/ f °f

(5)

and, on passing to the absolute values,

d -1 = df ° £-1f

In other words, inverse mappings have the same dilata

tion at corresponding points.

From (2) we obtain

(6) /l12 ° ££z /l12 0 f-/lf

lz 1 - 'ji.f/l12 0 f

If g is conformal, then /l12

(7)

o and we find

/l f .

If £ is conformal, /l f = 0 and

(8)f' 2/l o£ =(~

12 1£'[which can also be written as

(9)

In any case, the dilatation is invariant with respect to all con

formal transformations.

If we set g ° £ h we find from (6)

(10) £z /lh - /If

lz 1-'ji.f/lh

10

For the dilatation


(11)

and

d 0 fh 0 r 1

(12) log D -1 0 f == (Ilh' Il~ ,h 0 f

the non-euclidean distance (with respect to the metric ds

2 Idwl in Iwi < 1).l-lwl 2

We can obviously use sup (Ilh' Il~ as a distance be

tween the mappings f and h (the Teichmiiller distance). It

is a metric provided one identifies mappings that differ by a

conformal transformation.

The composite of a K 1-q.c. and a K2-q.c. mapping is

K1

K2

- q.c.

D. Extremal Length

Let r be a family of curves in the plane. Each y ( rshall be a countable union of open arcs, closed arcs or closed

curves, and every closed subarc shall be rectifiable. We shall

introduce a geometric quantity ,.\(r), called the extremal l€f1gth

of r, which is a sort of average minimal length. Its import

ance for our topic lies in the fact that it is invariant under con

formal mappings and quasi-invariant under q.c. mappings (the

latter means that it is multiplied by a bounded factor).

A function p, defined in the whole plane, will be called

allowable if it satisfies the following conditions:

Differentiable Quasiconformal Mappings 11

1. P? 0 and measurable.

2. A (p) = JJ p2 dx dy f, 0, 00 (the integral is over

the whole plane).

For such a p, set

Ly(p) f p IdzlY

if p is measurable on y*, Ly (p) 00 otherwise. We intro-

duce

L (p) = inf Ly (p)y~-r

and

Defini tion:

L (p)2sup -

p A(p)

for all allowable p.

We shall say that r 1 < r 2 if every y2 contains a y1

(the y 2 are fewer and longer).

Remark. Observe that r 1 C r 2 implies r 2 < r 1 !

THEOREM 2. If r 1 < r 2' then h(r 1) :S h(r2) .

Proof. If Y1 S; Y2 then

LY1

(p) < LY2

(p)

inf LY1

(p) < inf LY2

(p)

and it follows at once that h (r1) :S h(['2) .

* (as a function of arc-length)


Example 1. [' is the set of all arcs in a closed rec

tangle R which joins a pair of opposite sides.

For any pa

f p(x + iy) dx > L (p)

o

f f p dx dy > b L (p)

R

b2 L(p)2 < ab f f p2 dx dy < ab A(p)

R

L(p)2 < aA(p) - b

This proves A(n ~ a/b.

On the other hand, take p = 1 in R, P = 0 outside.

Then L(p) = a, A(p) = ab, hence A(n ;:: a/b. We have

provedab

Example 2. [' is the set of all arcs in an annulus

r1 < Iz I ~ r2 which join the boundary circles.

Computation:

ff p dr dO > 217 L (p )


f f p2 rdrde

Equality for p = IIr .

Example 3. The module of an annulus.

Let G be a doubly connected region in the finite plane

with C1 the bounded, C2 the unbounded component of the com

plement. We say the closed curve y in G separates C1 and

C2 if Y has non-zero winding number about the points of CI'

Let r be the family of closed curves in G which separate C1

and C2

• The module M(G) A(r)-I, Consider, for example,

theannulus G = Ir1 ::; Izi :s r21,

217

L (p) :s f0

217L (p)

< fr0

p de

r f f pdrdeL (p) log (r 2 ) <1

L(p )22 r

217 log (;:) f f / rdrd{)log (/) <1

L(p )2 217<

A(p) log( r2 / r1

)


so Up) = 1 and A (p)

M(G) il7 log(r2 /r1)·

Once again p = 1/217r gives equality. Indeed, for any y € 1

we have

1 < !n(y, 0)1 = .L [ Jdz 1 <.L f~ - L (y)- 217 Z - 217 IZ I - p ,y y

= il7 log(r2 /r1)' We conclude that

Suppose that all y € 1 are contained in a region nand let ¢ be a K -quasiconformal mapping of n on n '. Let

l' be the image set of 1.

THEOREM 3. K-1 .\(1) ~ .\(1') ~ K .\(1).

Proof. For a given p (z) define p'V;) = 0 outside

n' and

in n'. Then

fp'ldt:[ ?: f p [dz[

y y

f f ,2 dt:d = f f 2 [¢zl + [¢zl dxdy <_ KA(p) .p S TJ n P 1¢z[ - 1¢zl

This proves .\' ?: 1\1.\, and the other inequality follows by

con!:!idering the inverse.

COROLLARY. '\(1) is a conformal invariant.

There are two important composition principles.


*I. 11 + 12 = IY1 +Y21 Y1 f11 , Y2 f 12 1II. 1 1 u r 2

THEOREM 4.

a) A(11

+ 12

) > A(11

) + A(12

) ;

b) A(11

U 12

)-1 2: A(11)-1 + A(1

2)-1

if 11

, 12

lie in disjoint measurable sets.

Proof of a). We may assume that 0 < A(11

), A(12

) < 00

for otherwise the inequality is trivial.

We may normalize so that

L 1(p 1) A (p 1 )

L 2 (P2) = A (P2 )

Choose p

L(p) > L 1 (P1) + L 2(P2)

A (p) < A (P1 ) + A (P2)

L(p)2A = sup ATPT 2: A (P1 ) + A (P2 )

It follows that A 2: A 1 + A2 .

+

Proof of b). If A = A(11 U 12

) = 0 there is no

thing to prove. Consider an admissible p with L (p) > 0 and

set P1 = P on E1 , P2 = P on E2 , 0 outside (where E1

and E2 are complementary measurable sets with r 1 S E 1 '

* Y1 + Y2 means "Y 1 followed by Y2'"


['2 S E/ Then LI (PI) 2: L(p),

A(p) = A(PI) + A(P2)' Thus

A (p) A (PI)

L(p)2 > LI(PI)2 +

and hence

q.e.d.

E. A Symmetry Principle.

For any y let y be its reflection in the real axis,

and let y+ be obtained by reflecting the part below the real

axis and retaining the part above it (y U y = y+ U (y+)-).

The notations rand ['+ are self-explanatory.

THEOREM 5. If [' = r then ,\([') = Yi ,\(['+) .

Proof.

1. For a given P set p(z)

Then

max (p (z), P(z)) .

and

This makes

L+(p)2 ~ 2 L(p)2 S 2,\([')A(p) A(p)

and hence ,\ ([' +) S 2,\ ([' ) .

2. For given P set


in upper halfplane

in lower halfplane

Then

L +(p+) = L + ( +)_ (p) L _ (p)y y + y y+y

L (p) + L_(p) > 2L(p) .y y

On the other hand

2A (p)

and hence

L(p)2L +(p+)2

< Y2y

< Y2 ,\,(r+)A (p) A (p+)

,\,(r) < Y2 ,\,(r+) .

q.e.d.

F. Dirichlet Integrals

z

Let ¢ be a K-q.c. mapping from n to n '. The

Dirichlet integral of a C 1 function u «() is

For the composite u 0 ¢ we have


(U 0 c/J)z (ul;, 0 c/J)c/Jz + (ul;, ° c/J) ¢z

I(u ° c/J) zl $ <iul;,l ° c/J)<ic/Jzl + 1c/J:zP

and thus

(1) D(u0c/J) S KD(u) .

Dirichlet integrals are quasi~invariant.

There is another formulation of this. We may consider

merely corresponding Jordan regions with boundaries y;' y , .

Let v on y' and v 0 c/J on y be corresponding boundary

values. There is a minimum Dirichlet integral Do(v) for func

tions with boundary values v, attained for the harmonic func

tion with these boundary values. Clearly,

(2)

One may go a step further and assume that v is given

only on part of the boundary. For instance, if v == 0 and

v == 1 on disjoint boundary arcs we get a new proof of the quasi

invariance of the module.

In order to define the Dirichlet integral it is not neces

sary to assume that u is of class C 1. Suppose that u (z) is

continuous with compact support. Thus we can form the Fourier

transform

Differentiable Quasiconformal Mappings

Ii(~, Tf) = 2~ !! ei(X~+Y1])u(x, y)dxdy

nand we know that

(UX)A = - i~1i

(I).) A = - iTfIi

It follows by the Plancherel formula that

D(u) =! !(e+Tf2)[liI2 d~dTf

and this can be taken as definition of D (u) .

19

CHAPTER II

THE GENERAL DEFINITION

A. The Geometric Approach

All mappings ¢ will be topological and sense preserv

ing f rom a region 11 to a region 11'.

Definition A. ¢ is K-q.c. if the modules of quadrilat

erals are K -quasi-invariant.

A quadrilateral is a Jordan region Q, Q c 11, together

with a pair of disjoint closed arcs on the boundary (the b-arcs).

Its module m (Q) = a/b is determined by conformal mapping

on a rectangle

a

The conjugate Q* is the same Q with the complemen-

tary arcs (the a-arcs). Clearly, m(Q*) m(Q)-l.

21


The condition is m (Q ') ~ Km (Q ). It clearly implies

a double inequality,

l\lm(Q) ::; m(Q') S Km(Q) .

Trivial properties:

1. If ¢ is of class C1, the definition agrees with

the earlier.

2. ¢ and ¢-l are simultaneously K-q.c.

3. The class of K-q.c. mappings is invariant under

conformal mappings.

4. The composite of a K l-q.c. and a K2-q.c. map

ping is K1

K2

-q.c.

K-quasiconformality is a local property:

THEOREM 1. If ¢ is K-q.c. in a neighborhood of

every point, then it is K-q.c. in n.

Proof.

Q -.---.. -Q

We subdivide Q into vertical strips Qj and then each

image, Q.' into horizontal strips Q .. ' with respect to the ree-l IJ

tangular structure of Q.'. Set m .. = m(Q ..) etc. Then1 IJ IJ

m = ~ mj

.1. > ~J' _1_m

i- m

ij

~. _1_,J m ..

IJ

The General Definition 23

If the subdivision is sufficiently fine we shall have mI•J• < Km .~.

- IJ

This gives m $ Km'.

LEMMA. m = m1

+ m2

only if the dividing line is

x = m1.

Proof. Let the conformal mapping functions be f1

, f2.

Set p = If1 'I in Q1' P = If2 'I in Q2' P = 0 everywhere.

else. Then, integrating over Q

f f (p 2 - 1) dx dy 0

f f (p - 1) dx dy ::: 0

But fJ(p-1)2dxdy = ![[(p2-1)-2(p-1)]dxdy $ O. Hence

p = 1 a.e. and this is possible only if f 1 = f 2 = z.

THEOREM 2. A l-q.c. map is conformal.

Proof. We must have equality everywhere in the proof

of Theorem 1. This shows that the rectangular map is the iden-

tity.

B. The Analytic Definition

We shall say that a function u(x, y) is ACL (absolutely


continuous on lines) in the region n if for every closed rec

tangle R c n with sides parallel to the x and y-axes,

u(x, y) is absolutely continuous on a.e. horizontal and a.e.

vertical line in R. Such a function has of course, partial de-

rivatives ux' u y a.e. in n.The definition carries over to complex valued functions.

Definition B. A topological mapping ¢ of n ia K-q.c.

if

1) ¢ is ACL in n;

2) I¢zl ~ kJ¢zl a.e. (k = K-l)K+l .

We ;;hall prove that this definition is equivalent to the

geometric definition. It follows from B that ¢ is sense pre

serving.

We shall say that ¢ is differentiable at Zo (in the

sense of Darboux) if

¢(z)-¢(zo) = ¢z(zo)(z-zo)+¢z(zo)(z -zo)+ o(\z-zoP

The first lemma we shall prove is an amazing result due to Geh

ring and Lehto.

LEMMA 1. If ¢ is topological* and has partial deriva

tives a.e., then it is differentiable a.e.

By Egoroff's theorem the limits

¢(z+M-¢(z)h

P(z + ik) - P(z )k

'" The result holds as soon as ¢ is an open mapping.


are taken on uniformly except on a set n - E of arbitrarily

small measure. It will be sufficient to prove that ¢ is differ

entiable a.e. on E.

Remark. Usually Egoroff's theorem is formulated for

sequences. We obtain (1) if we apply it to

\¢(z+h)-¢(z) IsiP -¢)z) .

0< Ih < lin hThe set E is measurable, and therefore it intersects

a.e. horizontal line in a measurable set. On such a line a.e.

point in E is a point of linear density 1. The same holds for

vertical lines. Therefore a.e. xo' Yo ( E is a point of linear

density 1 for the intersections of E with x = Xo and y = Yo'

It will be sufficient to prove that ¢ is differentiable at such a

point Zo = Xo + iyo' and for simplicity we assume Zo = O.

Because of the uniformity ¢x and ¢y are continuous

on E. Given ( > 0 we can therefore find a [) > 0 such that

Ih I < [), [k I < [) and

(2)

as soon as

z ( E.

I¢)z) - ¢x(O)1 < ([¢/z) - ¢y(O)1 < (

I¢ (z + h) - ¢ (z )h

I¢ (z + ik) - ¢ (z)k

Ix I < [), [y I < [),

We follow the argument which is used when one proves

that a function with continuous partial derivatives is differen

tiable. It is based on the identity


¢(x+ iy) - ¢(Q) - x¢x(O) - Y¢y(O) =

[¢(x+ iy) - ¢(x) - y ¢y (x)] + [¢(x) - ¢(Q) - x ¢x (0)]

+ [y (¢y (x) - ¢y (0))]

If x f E we are able to derive by (2) that

The same reasoning can be repeated if y f E. Thus we have

proved what we wish to prove if either x f E or y f E.

We now want to make use of the fact that 0 is a point

of linear density 1. Let ml(x) be the measure of that part of

E which lies on the segment (-x, xL Then (ml(x))/2Ixl -> I

for x -> 0, and we can choose 8 so small that

m (x) > ~Ixl1 I + f

for Ix [ < 8. For such an x the interval (-IX,x) cannot be+f

free from points of E, for if it were we should have

Ix [ 2Hml(x) ::; Ixl +I+f = IH Ix[

The same reasoning applies on the y-axis. If [z I < _8_ weI +f

conclude that there are points xl' x 2' Yl' Y2 f E with

We may then conclude that (3) holds on the perimeter of the

* We are assuming for convenience that z lies in the first

quadrant.


rectangle (~, ~) x (Y1 ' Y2 ).

To complete the reasoning we use the fact that ¢ sat

isfies the maximum principle. There is a point z* = x* + iy*

on the perimeter such thai

I¢(x+ iy) - ¢(O) - x¢ (0) - y¢ (0)1x y

$ 1¢(x*+iy*)-¢(O)-x¢x(O)-Y¢y(O)1

< 3f Iz* 1+ Ix-x* 1 I¢ (0)1 + Iy-y* 1 1¢ (0)1x y

< 3dlH)lzl + fl¢ (O)llzl + fl¢ (O)llzlx y

This is an estimate of the desired form, and the lemma is proved.

We can say a little more. If E is a Borel set in n we

define A (E) as the area of its image. This defines a locally

finite additive measure, and according to a theorem of Lebesgue

such a measure has a symmetric derivative a.e., that is,

f(z) lim .AlQ)m(Q)

when Q is a square of center z whose side tends to zero.

Moreover,

f f(z)dxdy $ A(E)

E

(we cannot yet guarantee equality). But if ¢ is differentiable

at z it is immediate that f (z) is the Jacobian, and we have

proved that the Jacobian is locally integrable.

But f = I¢z 12- I¢Z' 12 and if 2) is fulfilled we ob-

tain


We conclude that the partial derivatives are locally square in

tegrable.

Moreover, if h is a test function (h ( C l with com

pact support) we find at once by integration over horizontal or

vertical lines and subsequent use of Fubini's theorem

(4)

f f ¢x h dxdy

f f ¢yh dxdy

-f f ¢hx dxdy

-f f ¢hydxdy

In other words, ¢x and ¢y are distributional derivatives of ¢.

More important still is the converse:

LEMMA 2. If ¢ has locally integrable distributional

derivatives, then ¢ is ACL.

Proof. We assume the existence of integrable functions

¢l' ¢2 such that

f f¢l h dxdy -f f ¢ hx dxdy

(5)

f f ¢2 h dxdy - f f¢hydXdY

for all test functions.

Consider a rectangle RTJ = {O ::; x ::; a, 0 ::; y ::; TJI and

choose h = h (x) k (y) with support in RTJ' In

f f ¢l h (x )k (y) dx dy = - f f ¢ h' (x)k (y) dx dy

RTJ RTJ

we can first let k tend boundedly to 1. This gives

The General Definition

f f ¢I h(x)dxdy

RTJand hence a

f ¢ 1(x, TJ)h (x )dx

o

f f ¢h'(x)dxdy

RTJ

a-f ¢(x, TJ)h '(x )dx

o

for almost all TJ. Now let h "" hn

test-functions such that °< h <- n-B- .! ). We obtain

n

run through a sequence of

1 and h "" 1 on (1.,n n

¢(a, TJ) - ¢(o, TJ)(6)

af ¢I(x,TJ)dx

ofor almost all TJ. The exceptional set does depend on a, but

we may conclude that (6) holds a.e. for all rational a. By con

tinuity it is then true for all a, and we have proved that

¢(x, TJ) is absolutely continuous for almost all TJ. Moreover,

we have ¢x "" ¢1., ¢y "" ~ a.e.

In other words, B is equivalent to B',

integrable distributional derivatives which satisfy

q.e.d.

¢ has locally

It is now easy to show that B is invariant under con

formal mapping. We prove, a little more generally:

LEMMA 3. If w is a C2 topological mapping and if

¢ has locally integrable distributional derivatives, so does

¢ 0 w , and they are given by


(6)

Proof. We note first that

(~x ~Y) and (X~ x1J)

1J x 1J y y~ Y1J

are reciprocal matrices (at correspon ding points). This makes

( x~ x1J) = ! ( 1Jy -~y)

Y~ Y1J J -1J x ~x

with J = ~ 1J - ~ 1J ' the Jacobian of w.x y y x

For any test function how we can put

II [(¢~ 0 w)~x + (¢1J 0 w)1J)(h 0 w)dxdy

I I [(¢~ 0 w) 7+ (¢1J 0 w) 1JjHh 0 w) Jdxdy

I I (¢~Y1J - ¢1JY~) h d~ d1J

I I ¢(- (hy1J)~ + (hY~) 1J) d~ d1J

I I ¢(-h~Y1J + h1JY~) d~d1J

I I (¢ ow)(-(h c 0 w) ~x - (h 0 w) 1J x ) J dxdys J 1J J

= I I (¢ 0 w)(-(h~ 0 w) ~x - (h1J 0 w) 1J) dx dy

= - I I(¢ ow)(h oW)x dxdy

q.e.d.

The last lemma enables us to prove:

B => A .


Indeed, in view of the lemma we need only prove that a

rectangle with modulus m is mapped on a quadrilateral with

m' ~ Km, and this proof is exactly the same as in the differen

tiable case.

We shall now concentrate on proving the converse:

A=>B.

First we prove: if ¢ is q.c. in the geometric sense,

then it is ACL.

Let A (77) be the image area under the mapping ¢ of

the rectangle a ~ x ~ f3, Yo ~ y ~ 77· Because A (77) is in

creasing the derivative A '(77) exists a.e., and we shall assume

that A '(0) exists.

In the figure, let Qi be rectangles with height 77 and

base b.. Let b.' be the length of the image of b.. We wishI I I

to show, first of all: If 77 is sufficiently small, then the length

of any curve in Qi' which joins the "vertical" sides is nearly

b. '.I

To do so we first determine a polygon so thatn

b ' - !..i 2


Next, take TJ so small that the variation of ¢ on vertical seg

ments is < €I4n. Draw the lines through 'k that correspond

to verticals. Any transversal must intersect all these lines.

This gives a length

n

> b.' - (1

1

On using the euclidean metric we have now, if ( <min ~ bi ',

b. ,2> 1

4A.1

b. ,21 <

4A i

b.K-l.

TJ

b. ,2::; ~_1_

b.1

~b. < 4K A(TJ) • (~b.)1 - TJ 1 *

But A (TJ}/TJ'" A '(0) < 00. This shows that ~ b/ ... 0 with

~ bi' and hence ¢ is absolutely continuous.

q.e.d.

Finally, if ¢ is K-q.c. in the geometric sense, it is

easy to prove that the inequality

(k = K-l)K+l

holds at all points where ¢ is differentiable. This proves

that A => B.

We have now proved the equivalence of the geometric

and analytic definitions, and we have two choices for the analy

tic definition.

* The proof needs an obvious modification if bi

' = 00 ••


COROLLARY 1. If the topological mapping ep satis

fies epz = 0 a.e., and if ep is either ACL or has integrable

distributional derivatives, then ep is conformal.

Observe the close relationship with Weyl's lemma.

COROLLARY 2. If ep is q.c. and epz = 0 a.e., then

¢ is conformal.

Finally, we shall prove:

THEOREM 3. Under a q.c. mapping the image area is

an absolutely continuous set function. This means that null sets

are mapped on null sets, and that the image area can always be

represented by

A (E) I IJ dxdy .

E

Proof. ep = u + iv can be approximated by C2 func

tions un + ivn in the sense thet un ~ u, vn ~ v and

Consider rectangles R such that u and v are abso

lutely continuous on all sides.

IfR

[(u ) (v) - (u·) (v ) 1dx dymx ny my nx f u dv

m n

JR


As m, n .... 00 the double integral tends to ffRJ dx dy. For

m .... 00 the line integral tends to

ju dVn j vn du

aR aRand for n .... 00 this tends to

j v du judv

aR aR(all because u and v, and therefore UV, are absolutely con

tinuous on aR).

We have proved

j j ] dxdy

R

for these R. It is not precisely trivial, but in any case fairly

easy to prove that the line integral does represent the image

area,'and that proves the theorem.

COROLLARY. ¢z f, 0 a.e.

Otherwise there would exist a set of positive measure

which is mapped on a nullset, and consideration of the inverse

mapping would lead to a contradiction.

Remark. It can now be concluded that Dirichlet integrals

are K-quasi-invariant under arbitrary K-q.c. mappings. It is

possible to prove the same for extremal lengths.

CHAPTER III

EXTREMAL GEOMETRIC PROPERTIES

A. Three Extremal Problems

Let G be a doubly connected region in the finite plane,

C1 the bounded and Cz the unbounded component of its comple

ment. We want to find the largest value of the module M (G)

under one of the following conditions:

I. (Grotzsch) C1 is the unit disk (121 $ 1) and

Cz contains the point R > 1.

II. (Teichmiiller) C1 contains 0 and -1; Cz con

tains a point at distance P from the origin.

III. (Mori) diam(C1 n Ilzl $ In ? A; Cz con

tains the origin.

We claim that the maximum of M (G) is obtained in the

following symmetric cases.

(). ~I.

35

R Go


II. -1 0 P

III. A~.

Fig.!.

Case l. Let r be the family of closed curves that se

parate C1

and C2

• We know that A(r) = M(G)-I.

Compare r with the family r of closed curves that lie

in the complement of C1

U IR I, have zero winding number

about R and nonzero winding number about the origin. Evident

ly, r c r, and hence A(r) ~ A(f). But r is a symmetric

family, and hence A(r) = 71 A(r +) by our symmetry principle.

SimilarlY, if r 0 is the family r in the alleged extremal case*,

then A(r0) = 71A(r0+) .

We show that r + = r 0+. Each curve y in r has

points PI' P2 on (-00, -1) and 0, R) respectively. Say they

divide y into arcs Yl and h so that y = Yl + Y2' Then- + - + - + (- , - + -)+ H - + - + - b 1y = y 1 + Y2 = Y1 + Y2 . ere y 1 + Y2 e ongs

to r 0' and we conclude that Y+ fro+. Thus t + C r 0+'

and the opposite inclusion is trivial.

We have proved A(r) ~ A(r) = A(r0)' hence M (G) <M(Go) .

q.e.d.

* We enlarge r 0 slightly by allowing it to contain curves which

contain segments on the edges of the cut (R, (0). This is, of

course, harmless.

Extremal Geometric Properties

Case II. Let z = f(() map $[ < 1 conformallyonto C1 U G

with [(0) = O. Koebe's one-quarter theorem gives jf'(0)I ~ 4P

with equality for G = G1 .Suppose f(a) -1. The distortion

theorem gives

1 = If(a) [ < [at 1['(0)[- (1- [a$2

< 4Pla[- (1- [a\)2

and equality is again attained when G = Gl' In other words

Ial ~ [a 11 where a1 belongs to the symmetric case. The

module M(G) equals the module between the image of C1 and

the unit circle.

Finally, by inversion and application of Case I, if [al is

given the module is largest for a line segment, and it increases

when la[ decreases. Hence G1 is extremal.

Case 1I/. Open up the plane by (= VZ. We get a figure

which is symmetric with respect to the origin

with two component images of C1 and two of C2 . The compo

sition laws tells us that M(G) ~ YiM(6') where 6' is the re

gion between C1- and C/. It is clear that equality holds in

the symmetric situation.


By assumption, C 1 contains points zl' z2' with

Iz 11 ::; 1, Iz2I ::; 1, Iz 1- z2I ~ '\. Let '1' '2 f C1+,

-'I' -, 2 f C1- be the corresponding points in the '-plane.

The linear transformation

w = '+ '1 '1 + '2,- '1 '1 - '2

carries (-'1' -'2) into (0, 1) and '1 .... 00, '2 .... W o where

w0= _ ( '2 + '1 )2'2- '1On setting u = ('2 + '1)/('2 -'1) one has

2('/ + '/)

'/ - '/

we get

lui __1_< £ J4-,\2

lui ,\

lui < 2 + J4-->.!.,\

Iwol < (2+~YOne verifies that equality holds for the symmetric case, and by

use of Case II it follows again that M (G) is a maximum for the

case in Figure 1 .

Extremal Geometric Properties 39

In order to confonn with the notations in Kiinzi's book *the extremal modules will be denoted

I. ~ log <1> (R)277

II.

III.

~ log 'P(p)277

~ log xC\.) .277

There are simple relations between these functions. Ob

viously, reflection of Go gives a twice as wide ring of type GI'

and one finds

(1)

Another relation is obtained by mapping the outside of the unit

circle on the outside of the segment (-1, 0). This gives

(2)

and together with (1), we find the identity

(3)

The previous computation in Case III gives, for instance,

(4)

*

x C\.) = <1> (V4+2X" + y4 - 2,\ ),\

Hans P. Kiinzi, Quasikonforme Abbildungen, in Ergebnisse der

Mathematik,Springer Verlag, Berlin, 1960.


(1)

B. Elliptic and Modular Functions

The elliptic integralz

w = fo

dz

maps the upper half of the normal region G1 on a rectangle

III

:8II

....._......J'O

with sidesp

a f dzy(z+ Uz(p-z)

0

(2) 00

b f dzy(z+ l)z(z-P)

P

Evidently,

(3)1

2" log 'I' (p) = 2~

This is an explicit expression, but it is not very convenient for

a study of asymptotic behavior. Anyway, we want to study the

connection with elliptic functions in much greater detail.

We recall that Weierstrass' ~-function is defined by

(4)


and satisfies the differential equation

where

41

(6)

(7)

It follows that the ek are distinct.

We set T = w 2 / wI and consider only the halfplane

1m T > O. In that halfplane

e3 -e1p(T) =

e2 -e1

is an analytic function .;, 0, 1. It is this function we wish to

study.

aT + bCT + d

P is invariant under modular transformations

with (: ~) == (~ ~) mod 2, for g;J does not

change and the wk

/2 change by full periods. The transforma

tion T' = T + 1 replaces p by 1/p, and T' = -1/T changes

pinto 1 - p. In other words

(8)peT + 1)

p( - ~ )

and these relations determine the behavior of p(T) under the

whole modular group.

For purely imaginary T the mapping by g;J is as indi-

cated:


~. _ $7////lII/l/1o C4)1

-2-

The image is similar to the normal region with P p/l-p,

and we find

(9) rCp) = .1. log qt (l-.-L)T7 -p

o < p < 1. It is evident from this formula that p varies mono

tonically from 0 to 1 when T goes from 0 to 00 along the

imaginary axis.

It can be seen by direct computation that p" 1 as soon

as 1m T .... 00, uniformly in the whole halfplane. If we make

use of the relations (8) it is readily seen that the correspondence

between T and p is as follows:

-:1 o :1


....... _ -t... ",- - I -

I

Fig. 2.

43

We shall let T(p) denote the branch of the inverse func

tion determined by this choice of the fundamental region. Ob-

serve the symmetries.

(10)

Relation (8) gives

(11)

Also

They yield

T(-I) ;: l+i2

T( 1-2i y'3 )+ 1 + i

T (t) ;: T (p) ±. 1

T(l-p) ;: - -+.npJ

rCp> ;: -T (p)

1 +iy'32

It is clear that eTTiT is analytic at p ;: 1 with a simple

zero. Therefore one can write

(12) p _ 1 - a e iTTT, (a > 0)

but the determination of the constant requires better knowledge

of the function ..

It will be of some importance to know how Ip -11 varies

when 1m T is kept fixed_ In other words, if we write T;: S + it


we would like to know the sign of a/as log Ip -11. This har

monic function is evidently zero on the lines s = 0, s = 1,

and a look at Figure 2 shows that it is positive on the half

circle from 0 to 1. This proves

is log Ip-ll > 0

in the right half of the fundamental region, provided that the max

imum principle is applicable. However, equation (12) shows

that a/as log Ip-ll .... 0 as r .... 00. Use of the relation (8)

shows that the same is true in the other corners, and therefore

the use of the maximum principle offers no difficulty.

We conclude that Ip -11 is smallest on the imaginary

axis and biggest on Re r = ± 1 .

The estimates of the function p (r) obtained by geomet

ric comparison are not strong enough, but classical explicit de

velopments are known. For the sake of completeness we shall

derive the most important formula.

The function~(z) - ~(u)

~(z) - ~(v)

has zeros at z = ± u + mWl + nW2

and poles at z = ± v + mwl + nW 2 •

A function with the same periods, zeros and poles is

00

IIn=-oo

F (z). (nw 2 ±v-z)

2m W1 _ e 1

(the product has one factor for each n and each ±). It is easy

to see that the product converges at both ends ..


2 17i .!!.±..zW

1 - e 1

217i y.±...zW

1 - e 1

F (z)

It will be convenient to use the notation q = el7iT

el7iW2 / WI. We separate the factor n = 0 and combine the

factors + n. This gives- 217i !!.::...Z

W1 _ e 1

217i Y.-zW

1 - e 1

00

ITn=l

+u+ Z217i - W-

1- q2n e 1

with one factor for each combination of signs.

It is clear that

so(z) - SO(u)

SO(z) - SO(v)

F(z)

F(O)

In order to compute

1 - P =e2 - e3e

3-e

1

we have to choose z = w2 /2, U = (WI + ( 2)/2. v = w/2

which gives

= q,

217iu

~e = - q,

Substitution gives

F(z) = 21+ q-l

00

IT1

(1 + ~n+2)(I+ ~n)2(I+q2n-2)

(1+ q2n+l)2 (l+q2n-l)2


(14)

and

F CO) =~ 1+ q-l (1 + q2n+l )2(1 + q2n-l)2II

2 2 (1 + q2n)4

00

(1 + q2n-l y1 n4q 1 1 + q2n

Finally00

( 1 + q'" )'(13) 1 - p = 16q II1 1 + q2n-l

A similar computation gives

00 (1 _q2n-l )8p = II

1 1 + q2n-l

«14) is obtained by the relation rCl/p) = rCp) ± 1 .)

We return to formula (9). It gives

(15) log 'PCP) = 17 1m T(I':P) = 17 1m T (1 + ~)

or

'PCP) qCp)-l for PpP+ 1

and, by (13),8

I-p 1 16q 00 (l+q2n )P+l ~ 1+ q2n-l

or

'PCP) 00

( 1 + q'" )'(16) 16 IIP+l 1 1 + q2n-l

(17)


This gives the basic inequality

W(p) S 16(P+ 1)

47

(18)

and

(19)

Because lI> (R )

lI>(R)-R-

W(R 2 - dh it follows further that

4 IT (_I_+_q,:..2_n__ ) 4

1 1+ q2n-l

lI>(R) < 4R

(20)

For X (A) we obtain

AX (A ) < 4( y'4+2X + y'4=2X)

AX(A) < 16 .

C. Mori's Theorem

Let , = ¢Az) be a K-q.c. mapping of Iz I < 1 onto

I( I < 1, normalized by ¢(O) = O.

Mori's theorem:

(1)

and 16 cannot be replaced by a smaller constant.

Remark. The theorem implies that ¢ satisfies a Holder

condition, and this was known earlier. It follows that ¢ has

a continuous extension to the closed disk Iz lSI. If we ap

ply the theorem to the inverse mapping we see that the exten

sion is a homeomorphism.

COROLLARY. Every q.c. mapping of a disk on a disk

can be extended to a homeomorphism of the closed disks.


In proving Mori's theorem we shall first assume that ephas an extension to the closed disk .. It will be quite easy to re

move this restriction.

For the proof we shall also need this lemma:

LEMMA. If ep (z) with ep (0) = 0 is K-q.c. and

homeomorphic on the boundary, then the extension obtained by

setting ep (1/ z) = 1/(¢;(z)) is a K-q.c. mapping.

Proof. It is clear that the extended mapping is K-q.c.

inside and outside the unit circle. It follows very easily that it

is ACL even on rectangles that overlap the unit circle. The K

q.c. follows.

Proof of (1). There is nothing to prove if IZ1 - z2 1 ~ 1,

say. We assume that IZ1 - z2 1 < 1.

Construct an annulus A whose inner circle has the seg

ment zl' z 2 for diameter and whose outer circle is a concentric

circle of radius Y2.

Case (i). A lies inside the unit circle..

Consider the mapping

w =


We obtain

1 log 1 < Klog ~ (111;-2~11;I;121)217 Iz2- z11 217

< K log 8217 1(2-1;1 1

49

This gives

Case (ii). A does not contain the origin. Now the image

of the inner continuum intersects 1(1 < 1 in a set with diam

eter ~ 1I; 1 - I; 21 , and the outer continuum contains the origin.

Hence

and we obtain

To rid ourselves of the hypothesis that ¢ is continuous

on Iz 1 = 1, consider the image region ~r of Iz 1 < r

--0


and map it conformally on !w! < 1 by t/Jr with t/J/O) = 0,

t/J/(0) > O. The function t/J/¢ (rz)) is K-q.c. and continuous

on Iz I = 1. Therefore,

1t/J/¢(rz2))-t/J/¢(rz1))1 < 16!z2- z1[1I K

Now we can let r .... 1. It is a simple matter to show that t/Jr

tends to the identity, and we obtain

< 161 z -z !l1K- 2 1

But this implies continuity on the boundary, and hence the strict

inequality is valid.

To show that 16 is the best constant we consider two

Mori regions (Case III) G (,\) and G ('\2). If they are mapped

on annuli we can map them on each other by stretching the radii

in constant ratio

Klog X ('\1)

log X ('\2)> 1

Because the unit circle is a line of symmetry it is clear that we

obtain a K-q.c. mapping which makes ,\ 1 correspond to '\2 .

We have now

/-- ....../ f'

I I \{ \

·,A1 1 I\ I I\ I /

" 1/....... __ /

/--]'/ "-I \I \I . A1 J\ /

\ /........... - '"


for small enough A2

. Thus

A2

? (A )1/K 16-f1 16 1/ K

51

and for large K the constant factor is arbitrarily close to 16.

There are of course strong consequences with regard to

normal and compact families of quasiconformal mappings.

THEOREM 1. The K~quasiconformalmappings of the

unit disk onto itself. normalized by ¢ (0) = 0, form a sequen

tiaily compact family with respect to uniform convergence.

Proof. By Mori's theorem the functions ¢ are equicon

tinuous. It follows by Ascoli's theorem that every infinite se

quence contains a uniformly convergent subsequence: ¢n -> ¢.

Because Mori's theorem can be applied to the inverse ¢n-1 it

follows that ¢ is schlicht. It is rather obviously K-q.c.

For conformal mappings it is customary to normalize by

¢(O) = 0, I¢ '(0)[ = 1. For q.c. mappings this does not make

sense: we must normalize at two points.

For an arbitrary region n we normalize by ¢ (a 1)

b1

, ¢(a2

) = b2

where of course a1

f. a2

, b1

f. b2

·

THEOREM 2. With this fixed normalization the K-q.c.

mappings in n satisfy a uniform Holder condition

11K1¢(z1)-¢(z2)1 ~ M[z1 -z21

on every compact set. The family of such mappings is sequenti

ally compact with respect to uniform convergence on compact sets.

52 Quasiconfotmal Mappings

Ptoof. For any z l' z2 f n there exists a simply con

nected region G en, not the whole plane, which contains z l'

z2' al' a2· If A is a compact set in n, then A x A can be

covered by a finite number of G x G. Hence it is sufficient to

prove the existence of M for G.

We map G and G' = ep (G) conformally on unit disks

as follows:

The compact set may be represented by Is I :::; to < 1. Mori's

theorem gives

(1 - Is I) :::; 16(1 -Iw 1)1/K

Hence Is I :::; to implies Iw I :::; Po (depending only on to)

and a similar reasoning gives f3 ~ fJ o .We know that

If we can show that (w) and s (z) satisfy uniform Lipschitz

conditions

we are through. These are familiar consequences of the distor

tion theorem.


For instance

53

Iz '(sl)1 > Colz'(O)1

laz - all < C'lz'(O)1

and hence

C'

The other inequality is easier.

It now follows that every sequence of mappings has a

convergent subsequence. To prove that the limit mapping is

schlicht we must reverse the inequalities. Although G' is now

a variable region this is possible because {3 < {3o' a number

that depends only on a (that is, on G, ai' az).

D. Quadruplets

Let (ai' az' a3, a4 ) and (b l , bz' b3 , b4 ) be two

ordered quadruples of distinct complex numbers. There exists

a conformal mapping of the whole extended plane which takes

ak

into bk

if and only if the cross-ratios are equal. If they

are not equal it is natural to consider the following problem.

Problem 1. For what K does there exist a K-q.c.

mapping which transforms one quadruple into another.


It was Teichmiiller who first pointed out that the problem

becomes more natural if one puts topological side conditions on

the mapping.

To illustrate, consider the following figure:

a4 aa

l.. 1..

a1 atThere exists a topological mapping of the extended

plane which takes the line 1. into 1.' and £ into £'. Obvi

ously, as a self-mapping of the sphere punctured at aI' a2 , a3,

a4

this mapping is quite different from the identity mapping (it

is not homotopic). Therefore, although the cross-ratios are

equal it makes sense to ask whether there exists a K-q.c.

mapping of this topological figure.

We shall have to make this much more precise. First of

all, there exist periods wI' w 2 such that, for T = w21wI,

P(T) =

Evidently, a linear transformation throws (aI' a2, a3, a4) into

(e 1, e2

, e3 , (0), and we may as well assume that this is the

original quadruple.

Let n be the finite '-plane punctured at e1, e2 , e3,

and let P be the z-plane with punctures at all points


m(w1 /2) + n (~ /2). The projection «oJ defines P as a cover

ing surface of n. In this situation we know that the fundamen

tal group F of n has a normal subgroup G which is isomor

phic to the fundamental group of P, and the group r of cover

transformations is isomorphic to F /G. We know F, G, and rexplicitly. F is a free group with three generators aI' a2 , a3 ,

representing loops around e 1, e2 , e3 . G is the least normal

subgroup that contains aI

2,a22,a

32 , (a

1a

2a

3)2, and r is

the group of all transformations z .... ± z + row 1 + nw2 . The

translation subgroup r 0 consists of all transformations z ....

z + mW I + nw2 . It is generated by ~ z = z + wI and A2 z =

z + w 2 .

* * *Consider now a second quadruple (e1 ' e2 ' e3 ' (0)

and use corresponding notations n*, F *, etc. A topological

* *mapping ¢: n .... n induces an isomorphism of F on F

*which maps G on G . This means that the covering ¢ 0 «oJ:

* * * *P .... n and «oJ: P .... n correspond to the same subgroup

G* of F*, so there is a topological mapping t/J: p .... p*

such that ¢ 0 «oJ = «oJ * 0 t/J. Clearly, Al * = t/J 0 Al 0 t/J-land A

2* = t/J 0 A

20 t/J-1 are generators of r 0*. We write

* * * * * *Al Z = Z + wI' A 2 Z = Z + w 2 and call (wI' w 2 ) the

base determined by t/J. Since ¢ does not uniquely determine

t/J, we must check the effect of a change of t/J on the base.

We may replace t/J by T* 0 t/J, T* I: r*, and we find that

the base (wI *, w 2*) either stays fixed or goes into (-wI *,

-W2*). In any case, the ratio r* = w 2*/ wI * is determined

*by ¢, and we say that two homeomorphisms of n on n are


equivalent if they determine the same T*. By methods which

are beyond our scope here, one can show that two maps of 11

on 11* are equivalent if and only if they are homotopic.

Problem 2. For what K does there exist a K-q.c. map

ping of 11 on 11* which is equivalent to a given CPo'

We need a preliminary calculation of extremal length. De

note by (y1 I the family of closed curves in 11 which lift to

arcs in P with endpoints z and z + CU l' In general, (my1 T

n y2 I denotes the family of curves which lift to arcs with end

points z and z + mcu 1 + ncu 2 .

LEMMA. The extremal length of (Y11 IS 211m T,

(T cu/cu 1).

Proof. We may assume that cu1

sider the segment e in the figure:

'T

/ 7-{/~----/(---,I

T. Con-

Its projection is a curve Y f (Y 11. For a given p in 11 set

p = p(&O(z))I&O'(z)l. We obtain

f p dx 2: L (p) ,

e


and we conclude that

A _2_$. 1m T

57

In the opposite direction, choose p so that p = 1. If

a curve y ( Iy 11 is lifted to P its image y leads from a point

z to z + 1. Hence the length of y is ~ 1, and we have

L (p) = 1, A (p) = Y2 1m T •

This completes the proof.

Of course, w 1 may be replaced by any cW 2 + dW 1

where (c, d) = 1. The result is that

(1) 2

1m (aT+ b )cT+d

where (~ ~) is a unimodular matrix of integers.

We are now in a position to solve Problem 2. We lift the

* *mapping ¢: U .... U to t/J: P .... P and choose the base

(w/, w/) determined by t/J. It is clear that the image under

¢ of Iy1/ is the class Iy 1*1 corresponding to w 1*. If ¢ is

K-q.c. it follows that

1\1 1m T $. 1m T* $. KIm T .

But !cY2 + dY11 is also mapped on ICY2* + dY1*1 and we

have also1m aT* + b

c? + d~ KIm aT+b

CT + d

for all unimodular transformations.

To interpret this result geometrically, let S be any uni

modular transformation and let U be an auxiliary linear trans-


formation which maps the unit disk Iw I < 1 on the upper half

planewith U(O) = r.

Mark the halfplanes bounded by horizontal lines through

Sr and K Sr.

We know that Sr* does not lie in the shaded part. Mapping

back by U-1 S-1 we see that U-1 r* does not lie in a

shaded circle

which is tangent to the unit circle at U-1 S-1 (00) and whose

radius depends only on K. But the points S-1 (00) are dense

on the real axis, and hence the U-1 S-1 (00) are dense on the

circle. This shows that r* is restricted to a smaller circle.

An invariant way of expressing this result is to say that the non

euclidean distance between rand r* is at most equal to that

between ib and iKb, or •

*d[r, r 1 < log K


THEOREM 3. There exists a K-q.c. mapping equivalent

*to ¢o if and only if d[T, T 1 ~ log K .

We have not yet proved the existence. But this is im

mediate, for we need only consider the affine mapping

(T* - 'T) Z + (T - T*) ZT ."

which is such that ljJ (-z) = -;fJ (z), ljJ (z + 1) = ljJ (z ) + 1,* .and ljJ (z + T) = ljJ (z ) + T . As such, ljJ covers a mapping ¢

*of n on n which is equivalent to ¢o' and the dilatation is

precisely ed [T, T*1 .

What about Problem I? Here it is assumed that ¢o

maps e l , e2 , e3 on e l *, e2 *, e3 * in this order. When does

¢ have the same property? Assume that ¢o is covered by

ljJo: P .... p* which determines the base (wI *, w 2 *) and that

¢ lifts to ljJ determining (CW2* + dw l *, aw2* + bw l *). It

is found that ¢ will map e l , e2, e3 on e l *, e2 *, e3 * if and

only if (~ ~) == (6 ~) mod 2 .

The set of linear transformations ~ ~~@ such that

(~ ~) is unimodular and congruent to the identity mod 2 is

the congruence subgroup (of level 2) of the modular group. The

solution to Problem 1 is therefore as follows:

THEOREM 4. There exists a K-q.c. of n on n*which preserves the order of the points e i if and only if the

non-euclidean distance from T to the nearest equivalent point

of T* under the congruence subgroup is < log K.


In connection with the lemma it is of some interest to

determine the linear transformation (~ ~) == (~ ~) mod 2 for

which the extremal length (1) is a minimum. We claim that this

is the case for the point T that lies in the fundamental region

IT ± Y21 < Y2, IRe TI < 1

Then

Ic T + dl 2: Ic Re T + d I 2: Id I - Ic Iand

ICT + dl

> 7'21 c I - II d I - Y21 c II = Id I or Ic I - Id I

Under the parity conditions Id I 2: 1 and either Id I - Ic I 2: 1

or I c 1 - Id I 2:. 1. It follows that Ic T + d 12 2:. 1 and thus A

is smallest for the identity transformation,

COROLLARY Let ¢ be any K-q.c, mapping of the

finite plane onto itself with K < V3. Then the vertices of any

equilateral triangle are mapped on the vertices of a triangle with

the same orientation.

Such a mapping can be approximated by piecewise affine

mappings. The triangle corres-

ponds to

l+iy'32

-1 + ivaand we know that the corresponding T = 2 ' The near-

t ' 'th I . -1 + i d h I'des pomt WI a rea p IS --2- , an t e non-euc 1 ean


*distance is log Y3. Hence K < y3 guarantees that Imp >0, and this means that the orientation is preserved.

The remarkable feature of this corollary is that it re

quires no normalization. The result is at the same time global

and local.

CHAPTER IV

BOUNDARY CORRESPONDENCE

A. The M-condition

We have shown that a q.c. mapping of a disk on itself

induces a topological mapping of the circumference. How regu

lar is this mapping? Can it be characterized by some simple

condition? The surprising thing is that it can.

Things become slightly simpler if we map the upper

halfplane on itself and assume that 00 corresponds to 00. The

boundary correspondence is then given by a continuous increas

ing real-valued function h(x) such that h(-oo) = - 00 and

h (+ 00) = + 00. What conditions must it satisfy?

We suppose first that there exists a K-q.c. mapping ¢

of the upper halfplane on itself with boundary values h (x ) .

It can immediately be extended by reflection to a K-q.c. mapping

of the whole plane, and we are therefore in a position to apply

the results of the precedin'g chapter. Namely, let e 1 < e3< e2

63


be real points which are mapped on e 1 ', e3 ', e2 '. If

p = p' =

and r, r' are the corresponding values on the imaginary axis

we have

K-1 1m r ~ 1m r' ~ Kim r

We shall use only the very simplest case where e 1, e~.

e2 , are equidistant points x - t, x, x + t and consequently p =

Yi, which corresponds to r = i. In this case we have that

Equivalently, this means that

or

(1) 1 - p UK) ~ p , ~ p UK)

Actually, what we prefer are bounds for

1 - p'-p

and from (1) we get

1 - p UK) h (x + t) - h (x) p UK)pUK) < h(x) -h(x-t} < I-p UK)

Even better, let us recall that p (r+ 1) = 1/p (r). For

this reason the lower bound can be written as p (1 + iK) - 1 and

by our product formula (Chapter III, B, (13)) we find

K 00 (1 -2nTTK ).8p(l+iK) - 1 = 16 e-TT II :;,.+~e _l_e-(2n-l)1TK

Boundary Correspondence

We have thus proved

) ( ) 1 h(x + t) - h (x ) ( )(2 MK- < h{x)-h(x-t) < MK

where

65

(3) () 1 rrK 00 ( 1- e-{2n-l)rrK ) 8MK =-e II

16 1 1 + e-2nrrK

We call (2) an M-condition. Obviously (3) gives the best possi

ble value, the upper bound

M(K) < 1~ err K

THEOREM 1. The boundary values of a K-q.c. mapping

satisfy the M(K) -condition (2).

It is important to study the consequences of an M-condi-

tion

(4) M-1 < h(x+t)-h(x)- h{x) - h{x + t)

< M

even quite apart from its importance for q.c. mappings. Let

H (M) denote the family of all such h. Observe that is is in

variant under linear transformations S: x ... ax + b both of the

dependent and the independent variables. In other words, if

h I: H(M) then Sl 0 h 0 S2 I: H(M). We shall let Ho(M)

denote the subset of functions h that are normalized by h (0)

= 0, h(1) 1.

For h I: Ho(M) we have immediately

(5) 1 < h (Y2) < MM+l M+l


and by induction this generalizes to

(6) 1 < <

This is true, with the inequality reversed, for negative n as

well. In other words we have for instance

which shows that the h f H0(M) are uniformly bounded on any

compact intervals (apply to h (x) and 1 - h (1-x)).

They are also equicontinuous. Indeed, for any fixed a

the functionh (a+ x) - h(a)h (a + 1) - h (a )

is normalized. Hence 0 s;. x s;. 1/2n gives

h(a+x)-h(a) < (h(a+l)-h(a))(MM)n+1

which proves the equicontinuity on compact sets. As a conse-

quence:

LEMMA 1. The space Ho(M) is compact (under uni

form convergence on compact sets).

Indeed, the limit function of a sequence must satisfy (4),

and this immediately makes it strictly increasing.

Actually, the compactness characterizes the H o(M) .

LEMMA 2. Let Ho be a set of normalized homeomor

phisms h which is compact and stable under composition with

linear mappings. Then Ho C Ho(M) for some M.

Boundary Correspondence 67

Proof. Set a = infh(-l) , (3 = suph(-l) for h f Ho'

There exists a sequence such that h (-1) .... a, and a subse-n

quence which converges to a homeomorphism. Therefore a >

-00, and the same reasoning gives (3 < O.

·For any h f Ho the mapping

k(x)_h(y+tx)-h(}) , t>O- h(y+ t} - My

is in Ho' Hence

<h (y- t) - h (y)

< {3ah (y + t) - h (y )

or1 <

h (y+ t) - h (y)< 1

a h(y) - h(y-t} -~

which is an M-condition.

We shall also need the following more specific informa-

tion:

LEMMA 3. If h f Ho(M) then

1

1 < f h (x) dx < MM+1 M+1

0

Proof. Let us set F (x) = sup h (x ), h f Ho (M) .

This is a curious function that seems very difficult to deter

mine explicitly. However, some estimates are easy to come by.

We have already proved that F <'/2) ::; M/(M + 1). Be-

causeh (tx)

hftT

6B Quasiconformal Mappings

we obtain, for x = Yz,h (1. )_2_ < F(Yz)h (t)

and hence

(7)

Similarly

for t > 0 .

h((l-t)x+ t) - h(t)

1 - h(t)

givesh(1+t) _ h(t)

21 - h(t)

For t < 1 this gives

h(l+t) < F(Yz) + (1 -F(Yz))h(t)2

and

(8) F(1+t) < F(Yz) + (l-F(Yz))F(t)2 -

Adding (7) and (8)

(9) F(~) + F( l~t.) < F(Yz) + F(t) .

Now1 2

f F(t)dt = Yz f F(~)dto 0

1

= Yz f (F (~) + F (~ + ~)) dt

o

1

< Yz F (Y2) + Yz ! F (t )dt

o


Hence1f F(t )dt :s F ('h)

o

and the assertion follows.

The opposite inequality follows on applying the result

to I-h(l-t).

Remark. From

M< M+l

the weaker inequalities

11

:s Jh dt :so

are immediate, and since they serve the same purpose, Lemma

3 is a luxury.

B. The Sufficiency of the M-condition.

We shall prove the converse:

THEOREM 2. Every mapping h which satisfies an M

condition is extendable to a K-q.c. mapping for a K that de

pends only on M.

12y

vex, y)

The proof is by an explicit construction. We shall in

deed set ¢(x, y) = u(x, y) + iv(x, y) wherey

u(x, y) = ~y J h(x+t)dt

-yy

f (h(x+t) -h(x-t))dt

o

(1)


It is clear that v (x, y) 2: 0 and tends to 0 for y .... O. More

over, u(x,O) = hex), as desired.

The formulas may be rewritten

X+Y

u = 1 f h (t )dt2y

(1) ,x~·y

x+y x

v 1 ( f h(t)dt - f h(t)dt)2y

x X-Y

and in this fonn it is evident that the partial derivatives exist

and are

Ux iy

(h(x+y) -h(x-y))

1 jX+Y hdt+ 21y (h(x+y) + h(x-y))-2i

X--Y

1- 2y2

X+Y

Jx

x

hdt - fX-Y

h dt) + iy

(h (x + y)

- h(x-y))

The following simplification is possible: if we replace

h (t) by hI(t) = h (at + b), a > 0, the M-condition remains

in force and ¢ (z) is replaced by ¢1(z) = ¢ (az + b). Thus

¢1 (i) = ¢ (ai + b), and since ai + b is arbitrary we need only

study the dilatation at the point i. Also, we are still free to

choose h(O) = 0, h(l) = 1.

The derivatives now become


U =x

U =y

'12(I-h (-1)),1

-'12 f h dt+ '12(1+h(-I)) ,

-1

v =x

v =y

'12 (1+ h (-1))1 0

-'12 f h dt - f h dt

o -1

The (small) dilatation is given by

d =\

( U - v ) + i (V + U y )x y x _

To simplify we are going to set

1

~ = 1- f h dt,

o

o

Tff3 = -h(-I) + f hdt

-1

ThusU x '12(1 + (3)

U y '!2 (~ - Tf(3)

giving

f3 -h(-I) ,

( > 0).

V x '12(1-{3)

v y '12(~+Tf{3)

d=«I-e') + f3(I-Tf)) + H(I+~) - f3 (I+Tf))

«1+e') + f3(I+~)) + H(I-e'} -f3(I-Tf))

1 + ~2 + f32 (1 + Tf2) - 2f3 (~+ Tf )

1+ ~2 + f32 (I+Tf2) + 2f3(~+Tf)


We have proved the estimates

.:s M, 1M+ 1 < (

1 < ...JLM+1 .:s." - M+1

(the last follows by symmetry).

This gives, for instance

1+d2

< M(M+ 1)1_d2

D < 2M(M+1)

< MM+1 '

As soon as we have this estimate we know that the

Jacobian is positive, from which it follows that the mapping ¢

is locally one to one.

It must further be shown that ¢ (z) .... 00 for z .... 00.

By (1) / we havex x+y

U1 (f hdt + f hdt)2y

x-y x

x+y x

v 2~ ( J hdt - J h dt )

x x-yx+y x

u2 + v22; [( f hdtY+ (f h dt)2 ]

x x-yy

If x :::: 0, u2 + v 2 1 ( J hdt Y> -2y20

0

If x .:s 0, u2 + v2 > 1 ( f hdt Y2?-y


and both tend to 00 for y ~ 00. When y is bounded it is

equally clear that u2 + v 2 ~ 00 with z.

Now ,= ¢ (z) defines the upper halfplane as a smooth

unlimited covering of itself. By the monodromy theorem it is a

homeomorphism.

1. Remark: Beurling and Ahlfors proved D < M2 . To

do so they had to introduce an extra parameter in the definition

of ¢.

2. Remark: It may be asked how regular a function

h (t) is that satisfies an M-condition. For a long time it was

believed that the boundary correspondence would always be ab

solutely continuous. But this is not so, for it is possible to con

struct functions h that satisfy the M-condition without being

absolutely continuous.

C. Quasi-isometry.

For conformal mappings of a halfplane onto itself non

euclidean distances are invariant. For q.c.-mappings, are they

quasi-invariant? Of course not. If noneuclidean distances are

multiplied by a bounded factor (in both directions) we shall say

that the mapping is quasi-isometric.

THEOREM 3. The mapping ¢ constructed in B 1S

quasi-isometric. Indeed, it satisfies

with a constant A that depends only on M.

74

(3)


It is sufficient to prove (2) infinitesimally; that is,

A-I ..hk.L < 1M..l < A..hk.Ly - v - y

Again, affine mappings of the z-plane are of no impor

tance, and it is therefore sufficient to consider the point (0, 1).

We have1 0

v (i ) Y:z f h dt - f h dt

o -1

and the estimates

M"2

are immediate (using Lemma 3).

This is to be combined with

and

l<,bzl2 = J/s [(l+e) + f32(1+r?) + 2f3(~+1])1

One finds that (2) holds with

A = 4M2(M+ 1)

We omit the details.

D. Quasiconformal Reflection.

Consider now a K-q.c. mapping <,b of the whole plane

onto itself. The real axis is mapped on a simple curve L that

goes to 00 in both directions. Is it possible to characterize L

through geometric properties?*

* For instance, we know already that L has zero area.


First some general remarks. Suppose that L divides

the plane into nand n* corresponding to the upper halfplane

H and the lower halfplane H*. Let j denote the reflection

z .... z that interchanges Hand H*. Then ep ° j ° ep-l is a

sense-reversing K2_q.c. mapping which interchanges n, n*

and keeps L pointwise fixed. We say that L admits a K2 _

q-c- reflection.

Conversely, suppose that L admits a K-q.c. reflection

w. Let [ be a conformal mapping from H to n. Define

H

in H,

in H*.IF = [F=wo[oj

2It is clear that F is K-q.c. So we see that L admits a re-

flection if and only if it is the image of a line under a q.c.

mapping of the whole plane. Moreover, we are free to choose

this mapping so that it is conformal in one of the half-planes.

We say that the conformal mapping [ admits a K~q.c. extension

to the whole plane.

. . * *We can also consIder the conformal mappmg [ from H

to n*. The mapping j 0/*-1 ° w ° [ is a quasiconformal map

ping of H on itself. Its restriction to the x-axis is h (x) =

[*-1 ° [, and we know that it must satisfy an M-condition. Ob

serve that L determines h uniquely except for linear transfor

mation (h(x) can be replaced by Ah(ax+b) + B).

(1)

76 Quasiconlormal Mappings

*

On the other hand, suppose that h is given and satis

fies an M-condition. We know that there exists a q.c. mapping

with these boundary values: we make it a sense-reversing map

*ping l from H to H . We cannot yet prove it, but there exists

a mapping ¢ of the whole plane upon itself which is conformal

*in H and such that ¢ 0 l 0 j is conformal in H . (This con-

dition determines J1 in the whole plane, and Theorem 3 of Chap

ter V will assert that we can determine ¢ when J1¢ is given.)

Thus ¢ maps the real axis on a line L which in turn deter

mines h.

How unique is L? Suppose L 1 and L 2 admit q.c. re

flections WI and w2' and denote the corresponding conformal

* *mappings by 11> 11 , 12, 12 . Assume that they determine the

same h = 1/-10 11 = 12*-1 0 12, The mapping

g = l/2 0 11-1

in °1 ,* I *-1 ...... *12 OlIn u 1 '

*is conformal in 0 1 U 01 and continuous on L1

. Is it confor-

mal in the whole plane? To prove that this is so we show that

g is quasiconformal, for we know that a q.c. mapping which is

conformal a.e. is conformal.

Introduce F1

and F2 as in (1). Form

G = F2

- 1 0 I * 0 I *-1 0 F2 1 1

in H*. It reduces to identity on the real axis, and we set G =z in H. Then G is q.c. Hence F2 0 G 0 F 1-1 is quasi

conformal. It reduces to 12* 0 11 *-1 in °1* and to 12 /0 11- 1

in °1 ; that is, to g.

*

Boundary Correspondence n

We conclude that g is conformal. Hence 12 differs

from 11 only by a linear transformation, and L is essentially

unique.

There are two main problems:

Problem 1. To characterize L by geometric

properties.

*Problem 2. To characterize f (and f).

We shall solve Problem 1. I don't know how to solve Problem 2.

The characterization should be in analytic properties of the in

variant f"j f'.

We begin by showing that a K-q.c. reflection retains

many characteristics of an ordinary reflection.

·z~o . i!

*We shall set ( == (U «() and suppose that (

¢C"Z), (0 = ¢(zo) when Zo is real.

All numerical functions of K. alone, not always the

same, will be denoted by C (K).

LEMMA 1.

< C(K)


Proof. We note that

pZo - Z

satisfies Ip I == 1. For any such p there is a corresponding

T situated on the lines Re T == ± %, 1m T 2: %.

We conclude that there is aT' corresponding to p

*('0 - ')/('0 - ,) that is within n.e. distance log K of these

lines. This means that T' lies in a W-shaped region indicated

in our figure.

We note that the points ± 1 where p = "" are shielded

from the W-region. Also, p .... 1 as 1m T .... "" and p .... -1 at

the points ± '12, provided that they are approached within an

angle. Therefore p' is bounded by a constant C (K) and this

proves the lemma.

Remark. It is not necessary to investigate the behavior

of p (T) at T = ± %, for the true region to which T' is restrict

ed does not reach these points.

Let 8 (,) be the shortest euclidean distance from 'to

L.

LEMMA 2.

C (K )-1 < C(K)

Boundary Correspondence

The proof is trivial.

79

*The regions nand n carry their own noneuclidean

metrics defined by

A Id~1Idz I

y

for ~ = /(z). The reflection w induces a K-q.c. mapping of

*H on H, and we know that it can be changed to a C (K) quasi-

isometric mapping. It follows that we can replace w by a re

flection w' such that (at points t w '(~))

C(K)-l Ald~1 s A*Idt I < C(K) A Id~1

But it is elementary to estimate A(~) in terms of 8(~).

"--Y'I \I ~o. I\ I, ./

..... - ... o w

For this purpose map n conform ally on Iw I < 1 with

w(~o) = O. Schwarz's lemma gives

But the noneuclidean line element at the origin is 21 dw I. SO

A(~o) = 2Iw'(~o)1 < _2_- 8(~o)

In the other direction Koebe's distortion theorem gives

'I. 1• Iw'(~o)1

128 (~o)


Combining the results with Lemma 2 we conclude

LEMMA 3. If there exists a K-q.c. reflection across L,

then there is also a C (K )-q. c. reflection which is differentiable

and changes euclidean lengths at most by a factor C (K) .

This is a surprising result, for a priori one would expect

the stretching to satisfy only a Holder con·dition.

I I

L

Now consider three. points on L such that ~3 lies be

tween ~1' ~2' Now P = (zl - z3)/(zl - z2) is between 0

and 1 which means that T lies on the imaginary axis. There*.

fore T lies in an angle

1 * -12 arc tan K - :s arg T :s 17 - 2 arc tan K

It can also be chosen so that IRe T*1ed to the following region

.,.-- ....",

o

*:s 1, so T is restrict-

Again it is obvious that Ip 1 has a maximum C (K) and

we have proved:


THEOREM 4. If ~1' ~3' ~2 are any three points on L

such that ~3 separates ~1' ~2 then

It is more symmetric to write

and in this form the best value of C (K) can be computed. It

corresponds to the point e ia and can be computed explicitly.

E. The Reverse Inequality.

We shall prove that the condition in the last theorem is

not only necessary but also sufficient. In other words:

THEOREM 5. A necessary and sufficient condition for

L to admit a q.c. reflection is the existence of a constant C

such that

(1) < C

for any three points on L such that ~ 3 is between ~1 and ~2'

More precisely, if K is given C depends only on K,

and if C is given there is a K-q.c. reflection where K depends

only on C.


Introduce notations by this figure:

lSI

Let Ak be the extremal distance * from ak to 13k in

0, and A/ the corresponding distance in 0*. Thus A1A2

* *= 1, A1 A2 = 1. Through the conformal mapping of 0,

let '1' '3' '2 correspond to x - t, x, x + t. This means that

Al = A2 = 1. Through the conformal mapping of 0*, '1' '3''2 correspond to h (x - t), h (x), h(x + t). If we can show

*that Al is bounded it follows at once that h satisfies an M-

condition, and hence that a q.c. reflection exists.

(2)

We show first that

c-2 e-2TT ::;

Al = 1 implies

I'2 -'11I '3 -'21

Indeed, it follows from (1) that the points of {32 are at distance

~ C-1 1'2 - '1 I from '2' while the points of a2 have dis

tance ::; C 1'3 - '2 I from '2· If the upper bound in (2) did

not hold, a2

and /32

would be separated by a circular annu

lus whose radii have the ratio e2TT• In such an annulus the ex-

* The extremal distance from a to {3 in E is the extremal length

of the family of arcs in E joining a to /3.


tremal distance between the circles is 1, and the comparison princi

ple for extremal lengths would yield ,\ 2 > 1, contrary to hypothesis.

This proves the upper bound, and we get the lower bound by inter

changing ~l and ~3 °

Consider points ~ ( a2

, ~/ ( f32

° By repeated application

of (1)

and with the help of (2) we conclude that the shortest distance between

a2

and f32 is :::: c4 e-2" 1~2 - ~31 . To simplify notations write

Because of (1), all points on ~ are within distance ~ from ~ .

Let r* be the family of all arcs in n* joining a2 and f320

Then

where p is any allowable function (recall Chapter I, D). We choose

p = 1 in the disk { I~ - ~21 ~ M1 + M2}, p = 0 outside that

.) *dISk. Then Ly(p ~ M2

for all curves y ( r , whether y stays

within the disk or not. We conclude that

'\2*~ 1 ( M2 )2

" M1 +M2

* *Since \ '\2

is proved.

1 , \ * has a finite upper bound, and the theorem

CHAPTER V

THE MAPPING THEOREM

A. Two Integral Operators.

Our aim is to prove the existence of q.c. mappings f with a

given complex dilatation Ill" In other words, we are looking for solu

tions of the Beltrami equation

(1) f-Z == !J.fz '

where Il is a measurable function and III I ~ k < 1 a.e. The so

lution f is to be topological, and f-Z' fz shall be locally integrable

distributional derivatives. We recall from Chapter II that they are then

also locally square integrable. It will turn out that they are in fact

locally LP for a p > 2.

The operator P acts on functions h f LP, P > 2, (with

respect to the whole plane) and it is defined by

(2) Ph«,) == _1 f f h(z)(~ - 1) dxdy" Z-s z

(All integrals are over the whole plane.)

LEMMA 1. Ph is continuous and satisfies a rmiforrn Holder

condition with exponent 1- 2/p.

85

86 Quasiconforma.l Mappings

The integral (2) is convergent because h £ LP and

t;/'(z(z-()) £ LP, 1/p + 1/q = 1. Indeed, 1 < q < 2, and for

such an exponent Iz(z-()I-q converges at 0, (;6 0) and 00.

The Holder inequality yields, for ( ;6 0,

IPh«()1 s 1;1 IIh lip II z(z:,) IIq

A change of variable shows that

and we find

(3)

with a constant Kp that depends only on p. «3) is trivially fulfilled

if (= 0.)

We apply this result to hI(z) = h(z+ (1) . Since

we obtain

which is the assertion of the lemma.

The Mapping Theorem

The second operator, T, is initially defined only for functions

h ( C02 (C2 with compact support), namely as the Cauchy princi-

pal value

(5)

We shall prove:

lim( .... 0

1

17

LEMMA 2. For h ( Co2, Th exists and is of class Cl.

Furthermore,

(6)

and

(7)

(Ph)z

(Ph)z

f f ITh 12

dxdy

h

Th

Proof. We begin by verifying (6) under the weaker assump

tion h ( COl. This is clearly enough to guarantee that

(Ph)~ = - ;; f f z~z?; dxdy

(8)

(Ph), = - ;; f f z~z?; dxdy .

Let y( be a circle wi th center , and radius (. By use of

Stokes' formula we find


117 II h-

z_z, dxdy 1 II hz

217i i="[ dz dZ

and

= lim£ .... 0

1217i J h dzz-'

1 I I zh~, dxdy = 1 I I dh dz17 217i ~

lim [- 1 I hdi" 1 f f ~dZd~ ]217i z-' + 217i£.... 0Iz~1 >£( -,)

y£

= Th(')

We have proved (6).

Observe that (8) can be written in the form

(9)P(h z )P(h)

h - h (0)

Th - Th(O) .

Under the assumption h £ C02 we may apply (6) to hz, and

by use of the second equation (9) we find

(10)(Th )z

(Th)z

These relations show that Th £ C 1, Ph £ C2 .

The Mapping Theorem 89

Because h has compact support it is immediate from the

definitions that Ph == 0(1) and Th == O(lz \-2) as z ~ c>o.

We have now sufficient information to justify all steps in the cal

culation

== ii f f Ph(Ph)zz dzdi"

which proves the isometry.

ii f f (Ph )hz dz di"

f f Ihl 2dxdy

q.e.d.

The functions of class Co2 are dense in L 2 . For this

reason the isometry permits us to extend the operator T to all

of L2 , by continuity. Unfortunately, we cannot extend P in

the same way, for the integral becomes meaningless when h is

only in L 2 , and even if we use principal values the difficulties

are discouraging.

The key for solving the difficulty lies in a lemma of

Zygmund and Calderon, to the effect that the isometry relation

(7) can be replaced by

(11)

for any p > 1, with the additional information that Cp ~ 1 for

p ~ 2. Naturally, this enables us to extend T to LP, and for

p > 2 the P-transform is well defined.


The proof of the Zygmund-Calderon inequality is given in

section D of this chapter. At present we prove:

LEMMA 3. For h ( LP, P > 2, the relations

(Ph)-Z h(12)

(Ph)z Th

hold in the distributional sense.

We have to show that

f !(Ph)¢-Z -! ! ¢h

(13)

! ! (Ph)¢z -! ! ¢Th

for all test functions ¢ ( Co 1. We know that the equations hold

when h ( C02 . Suppose that we approximate h by hn ( C0

2

in the LP-sense. The right hand members have the right limits

since 11Th - Thn II P $ CP II h - hn II p' On the left hand side we

know by Lemma 1 that

IP(h-hn)1 $ Kp Ilh-hnll p Iz I1-

2/

p

and since ¢ has compact support the result holds.

B. Solution of the Mapping Problem.

We are interested in solving the equation

(1)

where 111111"" $ k < 1. To begin with we treat the case where

11 has compact support so that f will be analytic at "".


We shall use a fixed exponent p > 2 such that kCp < 1.

THEOREM 1. If /l has compact support there exists a

unique solution I 01 (1) such that 1(0) = 0 and Iz - 1 ( LP.

Prool. We begin by proving the uniqueness. This proof

will also suggest the existence proof.

Let I be a solution. Then Iz = /lIz is of class LP,

and we can form PUz). The function

F = I-PU-)z

satisfies F z = 0 in the distributional sense. Therefore F

is analytic (Weyl's lemma). The condition I z - 1 ( LP implies

F' - 1 (LP, and this is possible only for F' = 1, F = z+ a.

The normalization at the origin gives a = 0, and we have

(2)

It follows that

(3)

If g is another solution we get

I z - gz = T[/lUz - g)l

and by Zygmund-Calderon we would get

and since k CP < 1 we must have Iz = gz a.e. Because of

the Beltrami equation we have also Iz = gz· Hence 1-g and

7 - i are both analytic. The difference must be a constant,

and the normalization shows that I = g.

92

(4)


To prove the existence we study the equation

h = T (/l h) + T /l .

The linear operator h ~ T (/l h) on LP has norm < k C < 1.- P

Therefore the series

(5)

converges in LP. It is obviously a solution of (4).

If h is given by (5) we find that

(6)

is the desired solution of the Beltrami equation. In the first

place /l (h + 1) ( LP (this is where we use the fact that /l

has compact support) so that P [/l (h + 1)1 is well defined and

continuous. Secondly, we obtain

(7)f-z /l(h+ 1)

T[/l(h+ 1)] + 1 h+l

and f z - 1 = h ( LP.

The function f will be called the normal solution of

(1).

We collect some estimates. From (4)

so

(8)

and by (7)


(9)

The Holder condition gives, from (3),

K(10) 1£((1) - £((2)1 s 1-~C II/lli p 1(1 -(211-2/p + 1(1 -(21

p

Let v be another Beltrami coefficient, still bounded by

k, and denote the corresponding normal solution by g. We ob

tain

f -g == T(lIf -vg )z z r z z

and hence

II f z -gz ll p s II T [v(fz- gz)lllp + lIT[(/l-v)fzll1p

< kCpllfz-gzllp + Cpll(/l-v)fzllp

We suppose that v .... /l a.e. (for instance through a se

quence) and that the supports are uniformly bounded. The fol

lowing conclusions can be made:

LEMMA 1. IIgz-fz lip .... 0 and g .... f, uniformly on

*compact sets.

More important, we want now to show that f has deriva

tives if ./l does. For this purpose we need first a slight gener

alization of Weyl's lemma:

* Since f-g is analytic at 00, the convergence is in fact uniform

in the entire plane.


LEMMA 2. If p and q are continuous and have locally

integrable distributional derivatives that satisfy Pz = qz'

then there exists a function f ( C 1 with fz p, f F = q.

It is sufficient to show that

f p dz + q dz 0

y

for any rectangle y. We use a smoothing operator. For ( > 0

define B/z) = 1/71(2 for [zl ::; (, B/z) = 0 for Izi > (.

The convolutions p * B( * B(, and g * B( * B(, are of

class C2 and

Therefore

and the result follows on letting ( and (' tend to O.

We apply the lemma to prove:

LEMMA 3. If 11 has a distributional derivative

o

p > 2,

then f ( C1 , and it is a topological mapping.

Proof. We try to determine ,\ so that the system

(11)

has a solution. The preceding lemma tells us that this will be

The Mapping Theorem

so if

95

(12)

or

,\-z (/1'\) z

(log '\)z

The equation

can be solved for q in LP, and we set

o = P (/1 q + /1z) + constant

in such a way that 0 ~ 0 for z ~ 00. Thus 0 is continuous

and

q

Hence ,\ = eO satisfies (12), and (11) can be solved with I

(C 1 . We may of course normalize by 1(0) = 0, and then I

is the normal solution since 0 ~ 0, ,\ ~ 1, I z ~ 1 at 00.

TheJacobian I/zI2-1/Z-12 = Cl-I/11 2)e20 is strictly

positive. Hence the mapping is locally one-one, and since f(z)

~ 00 for z ~ 00 it is a homeomorphism.

Remark. Doubts may be raised whether (eO) z = eO0 z

in the distributional sense. They are dissolved by remarking

that we can approximate 0 by smooth functions on in the sense

that on ~ 0 a.e. and (on) z ~ 0 z in LP (locally). Then for

any test function 1>,

96 Quasicon£urmal Mappings

q.e.d.

Under the hypothesis of Lemma 3 the inverse function

£-1 is again K-q.c. with a complex dilatation ii 1 = f1-1f

which satisfies lii1 0 £1 = 1111.

Let us estimate II iiI lip . We have

f fl'j11lPd~dTJ = f fIIlIP(I£zI2 -1£zI2)dxdY

and thus

If we apply the estimate (10) to the inverse function we

find

(13)

The Mapping Theorem

It is now obvious how to prove

97

THEOREM 2 For any Il with compact support and

IIIl to ~ k < 1 the normal solution 01 the Beltrami equation is

a q.c. homeomorphism with III = Il·

We can find a sequence of functions Iln £. C1 with

Iln .... Il a.e., Illnl ~ k and Iln = 0 outside a fixed circle.

The normal solutions In satisfy (13) with In in the place of

I and Iln in the place of Il· Because In.... I and II Iln \I p ....

1I1l11 p ' I satisfies (13), and hence I is one-one.

We recall the results from Chapter II. Because I is a

uniform limit of K-q.c. mappings In' it is itself K-q.c. As

such it has locally integrable partial derivatives, and these par

tial derivatives are also distributional derivatives. We know

further that Iz f 0 a.e. so that III = Iz/Iz is defined a.e.

and coincides with Il.

(Incidentally, the fact that I maps null sets on null

sets can be proven more easily than before. Let e be an open

set of finite measure. Then

mes In<e) = f <lln,zI2 -lln,z 12) dx dy

e

e

1 2/p(mes e) -


and since II f II is bounded we conclude that f is indeedn,z p

absolutely continuous in the sense of area.)

We shall now get rid of the assumption that /1 has com

pact support.

THEOREM 3. For any measurable /1 with 11/11100

< 1

there exists a unique normalized q. c. mapping f /1 with complex

dilatation /1 that leaves 0, 1, 00 fixed.

Proof. 1) If /1 has compact support we need only nor

malize f.

2) Suppose /1 = 0 in a neighborhood of O. Set

1 z2ii.(z) = /1 ('z) ~

z

Then ii has compact support. We claim that

Indeed,

f/1(z) = 1

f/1(11 z )

f/1(z)1 1

fii 01z)Z f ii(11z )2 7 Z

f!(z) 1 1 f!::. (IIz)Z fj1(l/z )2 _2 zz

Remark: The computation is legitimate because f ii is

differentiable a.e.

3) In the general case, set /1 = /11 + /12' /11 == 0

near 00, /12 = 0 near O. We want to find A so that

fA 0 /2 = f/1, fA == f/1 0 (1/12)-1 •

The Mapping Theorem

In Chapter I we showed that this is so for

and this ,\ has compact support. The problem is solved.

99

THEOREM 4. There exists a Il-conformal mapping of the upper

halfplane on itself with 0, 1, 00 as fixed points.

Extend the definition of Il by

;(z) = 1l(Z)

One verifies by the uniqueness that fll(i:) 71l(z). Therefore

the real axis is mapped on itself, and so is the upper halfplane.

SOROLLARY. Every q.c. mapping can be written as a

finite composite of q.c. mappings with dilatation arbitrarily close

to 1.

We may as well assume that f = fll is given as a q.c.

mapping of the whole plane. For any z, divide the noneuclidean

geodesic between 0 and Il (z) in n equal parts, the succes

sive end points being Ilk (z). Set

Then

(Ilk+1 - Ilk

1 - Ilk+1 Ilkf -1

o k


FIl/FIl(I) and FIl(I) ~ 1, the assertion

If fll is K-q.c. it is clear that gk = f k+1 0 fk-1 is

K l/n_q.c ., and

C. Dependence on Parameters.

In what follows fll will always denote the solution of

the Beltrami equation with fixpoints at 0, 1, 00. We shall need

the following lemma:

LEMMA. If k = 111111 00 ~ 0 then Ilf~ - 111 1,p ~ 0

for all p.

Note: II II R ,p means the p-norm over Iz I ~ R:

Suppose first that 11 has compact support, and let F 11

be the normal solution obtained in Theorem 1. We know that

h = F 11 - 1 is obtained fromz

h = T (11 h) + Til

and this implies Ilh lip ~ c 1111 lip ~ O. Here p is arbitrary,

for the condition k Cp < 1 is fulfilled as soon as k is suffi

ciently small.

Since f 11

follows for all 11 with compact support.

Let us write f(z) = 1/f ( ~ ). Then it is also true that

Iltll -11/ 1 ~ 0 when 11 has compact support. It is again suf-z ,p

ficient to prove the corresponding statement for the normal solu-

tion F 11. One has


_ 1 \P dxdy

Iz l4

For Iz I > R, F

The part ffl < Izl < R can be written

f f \z2(Fz(z) - 1) +

F (z)2 F (z)21<lzl<R

and tends to zero because IIF -l11 R .... O.z ,P

may be assumed analytic, and F (z) .... z uniformly.v

In the general situation we can write f = g 0 h where

J.Lh = J.Lt inside the unit disk and h is analytic outside. Then

J.Lh and J.L g are bounded by k and have compact support.

Sincev v v

f = (g 0 h)h + (g- 0 h)F = (g 0 h)hz z z z z z z

we get

Ilfz-1111,p < 1I[(iz -1) 0 h]hzlll,p + Ilhz - 1 11 1,p

For the first integral we obtain

f f l(gz-l) 0 hlPlhzl P dxdy

Izl<l

< _1_ f f Igz -liP Ih~ 0 h-1IP- 2 dx dy= 1-k2 _

hllzl<lI

~ 1_lk2 ( !! 1~_112Phllzl <11

. f f IhzI2P-4)~ 0 .

Izi < I


One of these integrals is taken over a region slightly bigger than

the unit disk, but this is of no importance. The lemma is proved.

It is now assumed that /l depends on a real or complex

parameter t and that

/l(z, t) = tv(z) + te(z, t}

where v and E f L"" and II E (z, t )11"" ... 0 for t ... O. We

shall show that f/l = f(z, t) has a t-derivative at t = O.

For any I~ I < 1 we can write

f(~) = 1 I f(z )dz217i z - ~

Iz 1=1- 1 II fz(z) dxdy

17 z - ~Izi < 1

(the Pompeiu formula). Replace z by 1/z in the line integral.

It becomes

1217i I

Izl = 1

f(1jz)dzz(l-z()

v

f(z )-1

1- z~dz

The convergence is guaranteed as soon as t is suffi-...ciently small to make K < 2. Indeed, the inverse of f satis-

fies a Holder condition with exponent 11K. For small Iz I wev

have thus, If(z)! > mlzl K and hence

IIzl=o

If(z)I-1 jzlldzl

11 - z~1


The constants can be recovered from the normalization

f(O) = 0, f(l) = 1. All told we get

(~ 2 z ~ z ) dx dy .1- z~ 1-z

f(~)

v

_1 f f ~z(z)17 f (z )2

Izl < 1

In the first integral, set

~ ~-1)- - + - dxdyz-l z

f- = Il fz = Il (fz-1) + Ilzv

~(z)and use a corresponding expression for f- withz(zl Z)2 Il (II z ) . Because

Hf -111 1 -->0' and ~ --> vz ,p t

we find

f(~) = lim HC) -Ct--> 0 t

_1 f f v(z)( z:~ ~ +~-1 ) dxdy

17 z-I -z-

Izl < 1

~ ranges over a com-

.~(~ -z2 1-z~

-~ f f v(l/z)

[z[ < 1

The convergence is clearly uniform when

~z )1-z dx dy .

pact subset of I~I < 1.

Finally, if 1/z is taken as integration variable in the

second integral, it transforms to the same integrand as the first,

104

and we have

where


i(I;) - kJJv(z)R(z, 1;) dxdy

R (z, !;) 1~ -

I;z-1

1;-1+ -- =

z

By considering f(rz), one verifies easily that this formula is

valid for all 1;, and that the convergence is uniform on compact

sets.

For arbitrary to we assume

in the same sense as above, and we consider

where

and

a~ Hz, t) = i 0 fila

= - 1 II ( v <to) ~S:) 0 (lllo)-1 R (z, fila (I;))dx dyIT I-lila 1

2 f Z

= - kII v (t 0' z )(I~Or R (lila (z) , fila (1;)) dx dy

The Mapping Theorem

which is the general perturbation formula.

We may sum up our results as follows:

THEOREM S. Suppose

wherev(t), Il(t), e(s, t) (Co, 111l(t)11 00 < 1,

and II e(s, t)1100 -+ 0 as s -+ O.

Then

f Il(t+s) (,) = f J1(t) (,) + ti (" t) + o(s)

uniformly on compact sets, where

105

1(" t)

If v (t) depends continuously on" t (in the L00 sense)

then we can prove, moreover, that (a/at) fez, t) is a continuous

function of t. It is enough to prove this for t = 0, so we have

to show that

By inversion, the integral over the plane can again be written as

the sum of two integrals over Iz I ::;. 1, and they can be treated

in a similar manner. We con~ider only the first part.

The important thing is that the improper integral con-

verges uniformly; that is,

!! Iff(z)12 IR(fIl(z), fll(')1 dxdy < E

\z-'i <0zl < 1


in a uniform manner. Indeed, this integral is comparable to

JJ IR(z, f J1 «())1 dxdy .

fJ1[z: Iz-<I <81The region of integration can be replaced by a disk with center

fJ1«() and radius < 28, say. The value of the integral is then

uniformly of order 0 (8).

For the remaining part of the plane (with a fixed 8) the

difference

tends rather trivially to 0, and it is also clear that

since II f J1 - 111 -+ 0 and the other factor is bounded.p •

D. The Calder~n-ZygmundInequality.

We are going to prove that the operator

Th (0 = lim _.1 f f h (z ) dx dyf -+ 0 17 (z _ ()2

Iz-(I >falready defined for h f C0

2, can be extended to LP, p 2: 2,

so that

(1)

We prove first a one-dimensional analogue, due to Riesz.

The proof is taken from Zygmund, Trigonometric Series, first

edition (Warsaw, 1935).

LEMMA. For

The Mapping Theorem

f ( Co 1 on the real line, set

107

Hf(~)

00

pr.v. 1 f J.i.!l dx .17 x-~

-xl

Proof. Set00

U + IV ~ f f(x) dx.. ~

, = ~ + i.", ." > 0

The imaginary part V is the Poisson integral, and it is immedi

ate that

V(~) = f(~)

The real part is00

u(~, .,,) 1 f x-t f(x) dx17

(x- ~)2 + .,,2-xl

00

1 f f(g+x) - f(g.·.,x) x 2dx

17 x x2 +.,,20

-+ Hf(~)

We observe now that

~ lul P

~lvlP

~ IF IP =

as .,,-+0.

p(p-l) lulp-2(u 2 + u 2)x y

p(p-l)lvI P- 2 (u} + u/)p2 IF IP- 2 (u} + u/)

It follows that

~ (IF IP - -p- lu IP)p-l p2(JFIp-2 - luI P- 2)(u} + u/)

> 0 .

108 QuasiconformaI Mappings

We apply Stokes' formula to a large semicircle

"'--,/ ,

I \L J

and find easily

00

f (IF «()I P - -p- lu«(W') d( ~ 0 .p-l

It is easy to see that the integral goes to zero for Tf .... 00.

Hence00 00

-""'l -""'l

for fixed Tf > O. We observe that

(J IF IP q'()2/P= IIu2+ v211p/2 < Il u2 1lp/2 + Il v2 1lp/2 .

Thus

Lettir.g Tf .... 0 we get the desired inequality.


We continue the proof of the Caldero'n-Zygmund inequal

ity (1), following the method in Vekua, Generalized Analytic

Functions, Pergamon Press, 1962.

We define the operator

dxdy

zlzl

again as a principal value. On setting zit can be written

re i() we see that

T* f(') = 1 f (~ 1 f('+re i()) - f('-re i())dr )e-i() d()2" r

o 0

This implies

00 H'+rei()) - H'-reie-, ~lIIT*fll < 17 max 1 fP 2 ()

17 r0

The norm on the right does not change if we replace , by 'ei(),

and then the integral becomes Hf()(') where f()(z) = f(ze i()).

The norm is of course a 2-dimensional norm. But it is clear

that our estimate of the I-dimensional norm can be used, for we

obtain

AP Ilf II PP () P

110 QuasiconformaI Mappings

so that, finally, II r*f lip :::; ~ Apllf lip'*We extend r to LP by continuity. The proof of (1)

will now be completed by showing that rf = - r*r*f for

f f C02 . This, of course, allows us to extend r to LP also.

We have ..1- _1_ = __1_ . If f f C 1 we obtainaz Iz I 2z Iz I 0

r*f(') - 1 IIHz+')..1- _1_ dxdy17 az Iz I

1 I I f (z+') _1_ dxdy17 z Iz I

(2)

1 ..1- 11Hz) dx dy17 a, Iz-~

~ ~ Ilf(Z) (lz~'1 - lh-) dxdy .

For any test function ¢ it follows that

This remains true for f f LP, because the integral on the

right is absolutely convergent (compare the proof for Pf).

This means that (2) is valid in the distributional sense.

The Mapping Theorem

We can now write

T*T*[(w)

111

(One should check the behavior for z = 0 and z = w. The

integral blows up logarithmically, and the boundary integral

over a small circle tends to zero.) We have to compute

...d... limaz R ~ ""

The differentiation and limit can be interchanged, for it is quite

evident that

aaz

tends to zero, uniformly on compact sets. Moreover, we can re

place our expression by


lim ~ f f 1 1 d~ dTf. J f f de dT/Jz Iz-w! I~-wl - Jz 1~llz-~

I~-w! <R. !~! <Rfor the difference, which is an integral over two slivers, tends

uniformly to zero.

By an obvious change of variable the first integral be-

comes

J~ f f!~I < Rllz-wl

R/lz- wi 217

J f f drdOJzoo -I=l_-'=r-'-ei-O1--

_% R. (z-w)lz-w!

and it is obvious that the limit is _ 17z-w'

cond limit is - 171 z .

We now obtain

Similarly, the se-

T*T*t(w) = ~ [1 f f t(z )(_1 - 1) dXdY]Jw 17 z-w Z

= -~ Pf(w) = -Tf(w)Jw

This completes the proof of (1).

The fact that C ~ 1 as p ~ 2 is a consequence ofp

the

The Mapping Theorem

Riesz-Thorin convexity theorem:

The best constant Cp

-+is such that log Cp

function of 1/p .

113

is a convex

Proof. We consider 1\1

P2 =1 both ~ 2.a1' a2

Assume

II Tfl11/al < C1l1fll1/al

II Tf Il 1/ < C211£111/a2 a

2

If a = (1- t )a1 + t a2 the contention is that

II Tf 11 1/ a :s; C11

-t C/ II f 11 1/ a <0 < t :s; 1).

Conjugate exponents will correspond to a and a' with

a + a' = 1. Similar meanings for a1

', a2

'. We note first that

I! Tf 11 1/ a = s~p f Tf· g dxdy

where g ranges over all functions of norm one in L 1/a' Be

cause simple functions (measurable functions with a finite number

of values) with compact support are dense in every LP it is no

restriction to assume that f and g are such functions.

For such fixed f, g set

1 = / Tf· g dx dy

The idea of the proof is to make f and g depend analytically

on a complex variable ,. 1 will be a particular value of an ana

lytic function cI> <'), and we shall be able to estimate its modu

lus by use of the maximum principle.


For any complex " define

ill1F(') If I· a _f

~' If IG(') Igi a -L

Ig Iwhere a(') = (1-')a 1 + 'a2 and a(')' = 1- a('). Ob

serve that , enters as a parameter: F (,) and G(,) are func

tions of z. We agree that F (,) = 0, G(,) = ° whenever

f=O, g=O. Note that F(t) = f, G(t) = g. Weset

¢(,) =

F (,) is itself a simple function I. Fj X j which makes

TF (,) = I. F. TX.. Similarly G(,) = I. G. X. *, say, and weII} }

find

<I> (,) = '"' F. G./!T X. . X.* dxdy .L.Jl} 1 }

We see at once that it is an exponential- polynomial: <I> «()

I. aje \' with real \. Therefore <I> (,) is bounded if ~ =

Re ( stays bounded.

Consider now the special cases ~ = ° and ~ = 1. For

~ = ° we have Re a(~) = a1 and hence

Ifl a/

a

Igl a1'1a '

It follows that

IIF (,) 11 1/ a1

II G (,) 11 1/ a '1

1 .

The Mapping Theorem

For simplicity we may assume that II f 11 11a

ly a normalization).

We now get

11S

1 (this is mere-

I¢(~)I ::; IITF(~)1I1/a IIG(~)1I1/a' < C 1 .1 1

A symmetric reasoning shows that

I¢(~)I ::; C2

on ~ == 1. We now conclude that

log I¢(~)I - (1-~) log C 1 - ~ log C2 < 0

on the boundary of the strip 0 ::; ~ ::; 1. Since the lefthand num

ber is subharmonic the inequality holds in the whole strip, and

for ~ == t we obtain the desired result.

CHAPTER VI

..TEICHMULLER SPACES

A. Preliminaries

Let S be a Riemann surface whose universal covering-S is conformallY isomorphic to the upper haIfplane H. The

-cover transformations of S over S are represented by linear

transformations of H upon itself which form a discontinuous

subgroup r of the group n of all such transformations. We

can write

S r \ H (orbits)

and the canonical mapping

77: H -+ r\His a complex analytic projection of H on S.

Conjugate subgroups represent conformally equivalent

Riemann surfaces. Indeed, if ro == Bor B o- 1, B o ( n then

z -+ BOz maps orbits of r on orbits of r 0 (for BoAz

(Bo ABo -I) Boz ). Therefore Bo determines a one-one con

formal mapping of So == r o\ H on S.

Conversely, let there be given a topological mapping

g: So -+ S .-

It can be lifted to a topological mapping g: So -+ S which

117

118

obviously satisfies

QuasiconformaI Mappings

17 0 g g 0 170

and we hav.e g

·,I-g-lSo --.;:;..g • S

If g is conformal, so is g

10

= B0

1 Bo-1 .

The classes of conform ally equivalent Riemann surfaces

correspond to classes of conjugate discontinuous subgroups of

n (without elliptic fixpoints).

But even if g is not conformal, it is still true that

A = g 0 Ao 0 g-1 ( 1

whenever Ao ( 1 0 , for

17 0 A 17 0 g 0 Ao 0 g-1g 0 170 0 Ao 0 g-1g 0 17

00 g-1

17

In other words, g defines an isomorphism () such that

() - --1AO = g 0 AO 0 g

It is not quite unique, for we may replace g by Bog 0 Bowhere B ( 1, Bo ( 1

0, This changes () into ()' with

Ao()' B 0 i 0 (Bo Ao Bo- 1) 0 g 0 B-1

which means that we compose () with inner automorphisms of

10

and 1. We say that () and ()' are equivalent isomorphisms.

Teichmiiller 5paces 119

LEMMA. 81 and 82

determine equivalent isomorphisms

01 and 02 if and only if they are homotopic.

Proof. If 8l' 82 are homotopic they can be deformed

into each other via 8 (t), say, which depends continuously on t.

We can then find g(t) so that it varies continuously with t,

and since

has values in a discrete set it must actually be constant.

Conversely, suppose 81' 82 determine. equivalent 01'

°2 , By changing i 1, i 2 we may suppose that 01 = 02' and

hence

Define g (t, z) as the point which divides the noneuclidean

line segment between i1

(t, z) and i2(t, z) in the ratio

t: (I-t).

Becausei

1(Az) Ail(z)

(A A 0)i

2(Az) Ag

2(z) 0

it follows thati(t, Az) Ai(t, z)

) -) 1Hence 8(t = 17 0 g(t 0170- is a mapping from 50 to 5,

and we have shown that 81 and 82 are homotopic.

Definition of T(50

) (the Teichmiiller space):

Consider all pairs (5, t) where 5 is a Riemann surface

and f is a sense-preserving q.c. mapping of 50 onto 5. We


say that (Sl' f 1) ...... (S2' f2) if f2 0 11- 1 is homotopic to a

conformal mapping of Sl on S2' The equivalence classes are

the points of T (So), and (SO' I) is called the initial point of

T (So),-

Every f determines a q.c. mapping I of H on itself,

and thereby an isomorphism () of ro ' Two isomorphisms corre

spond to the same Teichmiiller point if and only if they differ by

an inner automorphism of n.The space T (So) has a natural Teichmiiller metric: the

distance of (Sl' 11)' (S2' f2) is log K where K is the small

est maximal dilatation of a mapping homotopic to f2

0 f1

- 1

Let us compare T(So) and T(Sl)' Let g be a q.c.

mapping of So on Sl' The mapping

(S, f) .... (S, fog)

induces a mapping of T(Sl) onto T(So)' Indeed, if (S, f) '"

(S', f '), then (S, fog) '" (S', f' 0 g). This mapping is clear

ly isometric.

B. Beltrami Differentials.-

A q.c. mapping f: So .... S induces a mapping f of H

on itself which satisfies- -

(1) f 0 Ao = A 0 f

for A ':. Ao() Conversely, if f satisfies (1) it induces a map

ping f.

* For typographical simplicity both mappings will henceforth be de

noted by f.

Teichmiiller Spaces 121

From (1) we obtain

(A' 0 fH z

(A' 0 f)f-z

and thus the complex dilatation /If satisfies

/If = (/If 0 Ao)Ao 'I Ao '

or

(2)

A measurable and essentially bounded function /l which

satisfies (2) for all Ao (:ra is called a Beltrami differential

with respect to roo Another way to express the condition is to

say that

is invariant under r o '

Conversely, if /If does satisfy (2), then

/If 0 A = /Ifo

and it follows that f 0 Ao is an analytic function of f, or that

A = f 0 Ao 0 [I

is analytic, and hence a linear transformation.

The linear space of Beltrami differentials will be denoted

by B (r0) and its open unit ball with respect to the L 00 norm

is denoted by B1

<ro ) .

For every I! ( B 1 (ro ) we know that there exists a cor

responding f/l which maps H on itself. We normalize it so

that it leaves 0, 1, 00 fixed. It is then unique.

We set

122 Quasiconiormal Mapping<s

and write r 11 for the corresponding group, 011 for the isomor

phism. Since Oil represents a point in Teichmiiller space we

have actually defined a mapping

R1 era) .... T(Sa)

It is clearly continuous from L"" to the Teichmiiller metric.

There is an obvious equivalence relation: Ill"'" 112 if

and 0112 are equivalent isomorphisms. It is very diffi-

cult to recognize this equivalence by direct comparison of III

and 112. Because we cannot solve the global problem we shall

be content to solve the local problem for infinitesimal deforma

tions.

Before embarking on this road we shall discuss a differ

ent approach which has several advantages. The mapping ill

was obtained by extending 11 to the lower halfplane by symme

try. If instead we extend 11 to be identically zero in the lower

halfplane we get a new mapping that we shall call ill (again

normalized by fixpoints at 0, 1, ",,).

Clearly, ill gives a q.c. mapping of the upper halfplane

and a conformal mapping of the lower halfplane. The real axis

is mapped on a line L that permits a q.c. reflection.

J1 \ \ \

IIIIII \\\\\L


It is again true that

A = f 0 A 0 f -1Il Il a Il

is conformal, and hence a linear transformation (it is conformal

*in nand n , and it is q.c., hence conformal). We get a new

*group rll

which is discontinuous on nUn . We call it a

Fuchsoid group. We also get a surface r" n = S and a q.c.Il _ *

mapping So -> S as well as a conformal mapping So -> r \ n_ Ilwhere So has the conjugate complex structure of So .

Observe that f Il and fIl are defined for all Il ( L 00

with 111111 00 < 1 (Il ( B 1) even when the group ro reduces to

the trivial group. Our first result is

LEMMA 1. f Il = fV on the real axis if and only if

*fIl == fv on the real axis, and hence in H .

Proof 1). If f = f on the real axis if and only ifIl v

fIl (H) and fv (H) are the same and therefore

f 0 (l1l)-1 = f 0 (lV)-1 ,Il v

for both are normalized conformal mappings of H on the same

region.

2) Suppose fll = fV on the real axis. The mapping

h = (lV)-1 0 fll reduces to the identity on the real axis, so

h can be extended to a q.c. mapping of the whole plane by put

*ting h (z) = z in H . Consider the q.c. mapping A =f" 0 h 0 (I )-1. In f (H\ A = f 0 (I )-1 is conformal.

v Il Il v IlIn 9H),


is conformal. Thus A is a linear transformation, and the nor-

malization makes it the identity. This means fv fll

in H*.

q.e.d.

We shall now make the assumption that ro is of the

first kind, which means that it is not discontinuous at any point

on the real axis. It is then known that the orbits on the real

axis are dense. In particular the fixpoints are dense. Under

these conditions we can prove

LEMMA 2. III and 112 in B <ro) determine the same

Teichmiiller point if and only if f1 = fll2 on the real axis.

Proof. If fill = fll2 on the real axis, then A Il1 =

A Il2 on the real axis and hence identically. Therefore (/1 =(/2, and III and 112 determine the same Teichmiiller point.

Conversely, suppose (/1 and (J1l2 are equivalent iso

morphisms. This means there is a linear transformation S f nsuch that

A Il20 S = S 0 AIl1 for all A in ro .

We conclude that S maps the fixpoints of A Il1 on the

fixpoints of A Il2 (attractive on attractive). Since they corre

spond to each other we see that

S 0 fill = fll2

on the real axis. By the normalization, S must be the identity.

q.e.d.

COROLLARY. If ro is of the first kind, III and 112

determine the same Teichmiiller point if and only if

f f In H*.III 11 2


This means we may identify the Teichmiiller space T (So)

*with the space of conformal mappings of H of the form fll'

11 ( B1(ro) .

An even better characterization is by consideration ofthe

Schwarzian derivative

f ", (f ")2It ,zl = _11_ _ ~ L11 f' L. f'

11 11

Let us recall its properties with respect to composition. Con

sider

and let primes mean differentiation. We get

F '(z)

F"F'

F ",

IF, zl

For a better formulation, let us denote the Schwarzian by

[£1. The formula reads

[£ 0 g] = ([tJ 0 g )(g ')2 + [g]

There are two special cases. For f = A, a linear transforma-

tion,

[A 0 g] = [g]

and for g = A

[£ 0 A] = ([£] 0 A )(A') 2


On setting ¢J1 = [fJ11,

(¢ 0 A) A ' 2 [f 0

J1 J1[fJ11

We see that ¢J1 satisfies

(¢J1 0 A )(A ')2 = ¢J1

which makes it a quadratic differential. (¢ dz2 is invariant).

The following theorem is due to Nehari:

*LEMMA 3. If f is schlicht in the halfplane H, then

\[fll ~ ~ y-2

b bProof. Suppose that F (<:) - <: + 1 + 2 + '" is

~ 7schlicht for 1<:1 > 1. The integral ii II <:1 =r 'PdF measures

the area enclosed by the image of 1<: 1 = r, and is therefore po

sitive. One computes

F' = 1 - ~ + ...<:2

'PdF2~ f1<:1 =r

l- f (, + b1 + ... )( 1 - ~ - ... )d <:21 ;: <:2

= 17 ( r 2 - ~ _ ... )r 2

It follows that Ib 11 ~ 1. (More economically, Ib 112 + 21 b 212

+ ... + n Ibnl2 + ... ~ 1, which is Bieberbach's FHichensatz.)

Note that

F"= 2b 1

<:36b

F'" = _ _ 1

t

+ ...

gives [F 1

Teichmiiller Spaces

6b1- - +... and hence

~4

127

lim 1,4[Fll $ 6, -+ 00

Set ~ = Uz = (z-zo)/(z-zo), Zo xo+ iyo'

Yo < O. Consider F (,) = f(U-l ,). Then

[E] = ([F] 0 U)U,2

Here

-2iyU' = __0_

(z-zo)2

2iyoU ,.., --

z-zo

U,2 '" _ ~ U 4 (z -+ zo) .4 Yo

On going to the limit we find

and then

q.e.d.

In view of the lemma it is natural to define a norm on

the quadratic differentials by

II¢> II = sup I¢> (z)1 y2

'128 Quasiconformal Mappings

C. ~ is Open.

We have defined a mapping /1 .... ¢/1 from the unit ball

B 1(r) to the space Q(r) of quadratic differentials with finite

norm. The image of B 1(r) under this mapping will be denoted

by ~ (r). It is our aim to show that ~ (r) is an open sub

set of Q(r).

The question makes sense even in the case where r is

the trivial group consisting only of the identity. In this case

the spaces will be denoted by B l' ~, Q.

THEOREM. ~ is an open subset of Q.

Clearly, ~ consists of the Schwarzians [f] of functions

f which are schlicht holomorphic in the lower halfplane and have

a q.c. extension to the upper halfplane. We know already that

all ¢ in ~ satisfy II¢ II s 312 .

LEMMA 1. Every holomorphic ¢ with II¢ II < Y:z 1S m

~.

Proof. We choose two linearly independent solutions of

the differential equation

(1) TI" = - Y:z ¢ TI

which we may normalize by TI~ Tl2 - TI; Till. It is easy to

check that f = TI/Tl2 satisfies [[1 = ¢. Observe that the

solutions of (1) have at most simple zeros. Hence f has at most

simple poles, and at other points f' f O.

We want to show that f is schlicht and has a q.c. exten

sion to the upper halfplane. To construct the extension we con

sider

TeichmiiIIer Spaces 129

111(Z) + (Z-Z)11{(Z)

112(z) + (z - Z)11;(Z)*(z (. H )

We remark first that the numerator and denominator do not vanish

simultaneously (because 11{ 112 - 112 111 = 1). Therefore F is

defined everywhere, but could be 00.

(112 + (Z"- z) 11; )2

Y:z¢(z-z)2

F-z

Simple computations yield

1

(2)

and thus

= Y:z¢(z-z) 2

The assumption implies \Fz I ::; k \Fz I for some k < 1.

Hence F is q.c. but sense-reversing.

z ( H

= { f(z)F(z)

f (z)

The extension will be defined by

z (. H*(3)

It must be shown that f is a q.c. 1-1 mapping.

This is easy if ¢ is very regular. Let us assume, spe

cifically, that ¢ remains analytic on the real axis, and that it

has a zero of at least order 4 at 00. It is immediate that f and

F agree on the real axis, and it is also evident that ; is local

ly schlicht. The assumption at 00 means that there are solu

tions of (1) whose power series expansions at 00 begin with 1

and z respectively. We have thus

130

F( z) =


111 a1z + b1 + 0(\:1)

112 a2z + b2 + 0(1:1)1. This gives

a1z + ~ + oqz 1-1)

a2

z + b2

+ O(lz 1-1)

which is also the limit of f.

Now the schlichtness of f follows by the monodromy~

theorem. We may of course, compose f with a linear transfor-

mation to make it normalized.

To prove the general case we use approximation. Put

S z = (2nz-i)j(iz+2n). Then S H* CC H* and S z ~ Zn n nfor n ~ 00. Set cPn(z) = cP (Snz) Sn '(z)2. We have

y 2IcPn(z)1 = IcP(Snz )IISn '(z)1 2y 2 < IcP(Snz )I(ImSn(z))2

and hence IIcPn ll ~ IIcP II· Now cPn has all the regularity pro-~ ~ *

perties. We can find mappings f with [f 1 = -!.. in Handn n ~n

uniformly bounded dilatations. By compactness there exists a

subsequence which converges to a solution f of the originaln

problem.

If cPn ~ cP it is not hard to see that a normalized solu

tion of 11" = - J.h. cPn1l con verges to a normalized solution of

11" = - J.h. cP1I' Therefore, if we choose the same normalizations

- - *-we may conclude that fa = f in H and in H . Hence f can

be extended by continuity to the real axis and is a solution of the

problem with- 2

J1 = -2cPY


But if 1> l Q(r) one verifies that /1 l B (r ), and we

conclude

LEMMA 2. The origin of Q(r) is an interior point of

Suppose now that 1>0 l ~ and [fo1 = 1>0 where fo =* *f /1

0We assume that f0 maps H on n, H on n . Then

the boundary curve L of n admits a q.c. reflection A. Ac

cording to Lemma 3 of Chapter IV D, we can choose A so that

corresponding euclidean lengths have bounded ratio. This means

that A is C (K )-q.c. and

C(K)-1 < IAzl < C(K)

provided that fo is K-q.c.

If [f] 1> the composition rule for Schwarzians gives

1> - 1>0 = If, follo,2

The noneuclidean metric in n* is such that

p(Old(' =~-y

Therefore, 111> - 1>0 II S l shows that g

l[gl(OI S l p«()2

f 0 fo-1 satisfies

For sufficiently small l we have to show that g has a

q.c. extension.

We set t/J [g] and determine normalized solutions 71 1'


This time we construct

g«() = 171«()/172«(), ( ( 0*

i «()171«(*) + «(-(*) 17 1 '«(*)

(( 0 .172«(*) + «(-(*)172 '«(*)

(Recall that (* = A«(L) Computation gives

/1 ;( ()'h. «(- (*)2 t/J «(*) A ,«()

((01+ 'h. «(- (*)2 t/J «(*) A(()

* < Cp«(*)-1 .But IA(I < 1\::1 ~ C(K) and 1(-( I Hence

1/1;1 ~( . C(K) < 1(4) 1 - ( . C(K)

as soon as ( is small enough.A

It must again be shown that g is continuous and schlicht.

There is no difficulty if L is analytic and t/J is analytic on L

with a zero of order four at 00.

The general case again requires an approximation argu-

ment. Let f = f 0 S where S is as in the proof ofnOn' n

Lemma 1, and let L n be the image of the real axis under In'

Ln is an analytic curve that admits a K-q.c. reflection, and t/Jis analytic on Ln'

The Poincare density p of 0 * = f (H*) is > p,n n n -so that It/J I ~ (p2 implies It/J I ~ (pn 2 . Therefore we can

construct a sequence of normalized q.c. mappings i such thatA *[gn1 = t/J in On and /1; satisfies the inequality (4). A

A n A

subsequence of the gn tends to a q.c. limit g which is equal


to g in n*. This proves Theorem 1. Since /l; satisfies (4)

we conclude:

COROLLARY. For every sequence of ¢n ( ~ converg

¢o = [f/lo 1 ( ~, there exist /l n .... /lo such that

= ¢n'

The proof is by writing ¢ = ~ 0 f 01 .

We come now to the most delicate part:

THEOREM 2. ~ (r) is an open subset of Q(r) .

.Remark: This was first proved by Bers. The idea of the

proof that follows is due to Clifford Earle.

Given any /lo ( B, we construct a mapping

f3 0 : ~ .... ~

as follows: Given ¢ ( ~ there exists a /l ( B 1 such that

¢ = ¢/l' With this /l we determine A by

(5) fA = f/l 0 ([ /l0)-1

and hence ¢A = ¢A .1

It is evident that f30

is 1-1, and it carries

zero. Moreover, f30

is continuous. For if ¢n .... ¢

have just proved the existence of /In .... /l such that

The corresponding ¢A converge to ¢A'n

and set f3 0 (¢) = ¢A' It is unique, for if ¢ = ¢ , then f /l/l /l /l1

has the same boundary values as fl. Hence fA has the sameA

boundary values as f 1


LEMMA. (Earle). <P1l

( Q(f') if and only if for every

A (f' there exists a linear transformation B such that

fll 0 A 0 ([11)-1 = B on R.

Proof. <P1l

( Q([') is equivalent to [ill 0 A] = [ill]'

and this is true if and only if f11 0 A 0 f11-1 = t, a linear

*transformation, in nil .1) If B exists, then

C = f 0 A 0 f -1 = f 0 (f11)-1 0 B 0 f 11 0 ([ )-111 11 11 11*

on f 11 (R). The first expression is holomorphic in nil' the

second in nil' Hence C is a linear transformation. *2) If C exists, then

B = f 11 0 A 0 ([ 11)-1 = f 11 0 f -1 0 C 0 f 0 (f11)-111 11

on R. But the last expression is a conformal mapping of H on

itself, hence a linear transformation. The lemma is proved.

Assume now that 110 ( B 1(f'). We find that f3 0 maps

~ (f') on ~ (f'1l 0 ) , for

fA 0 Allo 0 (fA)-1 = fll 0 A 0 ([11)-1 = All

From the lemma we infer that Q(f') n ~ is mapped on

Q(f'1l0 ) n ~. The origin has a neighborhood N in Q(f'1l0 )

which is contained in ~ (f'1l0 ). Write N = Q(f'1l 0 ) n No =

Q(f'1l 0) n ~ n Nowhere No is a neighborhood in ~. Then

f3-1(N) = Q([') n ~ n f3-1(No) = Q(f') n f3-1(No)

which is a neighborhood of <P in Q(f'). From N C ~ (f'1l 0 )110

* Because it is quasiconformal and conformal a.e.

Teichmiiller Spaces

we get f3 -1 (N) c ~ (r) and this proves that ¢ has a/1 0

neighborhood in Q(r) which is contained in ~ (r) .

135

q.e.d.

Conclusion:

If the group r of cover transformations of H over So

is of the first kind, the Teichmuller space T (S0) is identified

with ~ (r), an open subset of Q(r). The Teichmiiller metric

defines the same topology as the norm in Q(r) .

Consider the case where So is a compact Riemann sur

face of genus g > 1. Then Q(r) has complex dimension

3g- 3. Choose a basis ¢t' ... , ¢3~-3' Every ¢ f Q(r) is a

linear combination

¢ = 71¢1 + ... + 73_-3 ¢3~-3

with complex coefficients. We find that ~ (r) can be identi

fied with a bounded open set in C3~-3.

We can show, moreover, that the parametrization by 7 =

(71 , ... , 73~-3) defines the Riemann surfaces of genus g as a

holomorphic family of Riemann surfaces.

According to Kodaira and Spencer a holomorphic family

may be described as follows:

There is given an (n+ I)-dimensional complex manifold

V and a holomorphic mapping 17: V .... M on an n-dimensional

complex manifold M. Each fiber 17-1(,), 7 f M is a Riemann

surface.

The complex structures of V and M are related in the

following way: There is an open covering lual of V so that

136 Quasiconiormal Mappings

for each a there is given a holomorphic homeomorphism ha:

Ua ~ C x M connected by ¢af3 = ha 0 hf3-1 (for intersect

int Ua' U(3)' For any T ( M the restriction of ¢af3 to

hf3 (Ua n Uf3 n 7T-1 (r)) shall be complex analytic (in fact, these

functions determine the complex structure of 7T-1(T).)

In our case, M will be the set ~ (r). Every T ( ~ (r)

*determines a ¢ = [i,), and ill is uniquely defined in H .

*Hence Oil' Oil' and the group r Il are determined by T. To

emphasize the dependence on T we change the notations to ¢T'

*0T' rT

, etc. Since iT is determined in H by the differential

equation [iT1 = ¢T' we know that iT depends holomorphically

on the parameter T. For A ( r, the corresponding AT ( r T

*is determined by the condition iT 0 A = AT 0 iT in H . This

means that AT depends holomorphically on T, a fact that will

be important.

The Riemann surfaces in our family will be S (T) = 0/rT

and V will be their union. Thus the points of V are orbits

r l with <: ( 0T and T ( ~ (r). The projection 7T: V ~ M is

defined so that 7T-1(r) = S(r).

Consider a point r T <:0 in V. We pick a fixed <:0 fromo

the orbit and d_etermine an open neighborhood N (<:0) such that

N is compact, contained in 0T • and does not meet its imageso

under r T' The neighborhood N (s <:0' TO) of r T <:0 will con-o 0

sist of all r T<: such that II ¢T - ¢T II < E and <: ( N (<:0) ._ 0

Here E shall be so small that N does not meet its images under

r T' This is possible because AT is near AT0 when ¢T is


near ¢T' As a consequence, there is only one (( N «(0) no

r T( and the mapping h: r l.... «(, T) is well-defined in

N (E, (0' TO) ,

The neighborhoods N (E, (0' TO) shall be a base for the

topology of V. We assert, in addition, that the parameter map

pings h: r l.... «(, T) defined on the basic neighborhoods make

17: V .... M into a holomorphic family, Indeed, if two basic neigh

borhoods U0 and U1 intersect, then in U0 n Ul' hO(rT() ==

«(, r) and h1(rT() == (AT (, T) for some AT ( r

T• Hence the

mapping h1 0 ho-1 is given by «(, r) .... (AT(, T), which we

know is holomorphic in both T and (. It is evident that the pa

rameter mappings define a complex structure on V which agrees

with the conformal structure of the surfaces S (T) and makes the

mapping 17: V .... M holomorphic.

D. The Infinitesimal Approach.

We continue with a direct investigation of f 11, All that

does not make use of the mappings fll

,

For any function F(p.) and any y ( L C>O we set

limt .... 0

when the limit exists, and we omit the argument 11 when the de

rivative is taken for 11 = O. We are assuming that t is real.

We have already derived the representation

i[vl«() == - ~ f f v(z)R(z, ()dxdy


where

R (z, ,) = 1 _ U - ~z-' z z-1

We apply the formula to the symmetric case: v(Z") = i/ (z) ,

and we write more explicitly

(1) i[v 1(') = - ; ff v(z)R(z, ')dxdy

H

-; f f i/(z)R(Z", ,) dxdy

H.It is evident that f[v] is linear in the real sense, but

not in the complex sense. To obtain a complex linear functional

we form

(2)

and we find

(3) lI>[v](') = - ~ ff i/(z)R(Z", ')dxdy

H

It is holomorphic for 'l H. Its third derivative is

17 ff i/(z) dxdy(Z" _,)4

H

If v l B (r) one verifies that ¢ is a quadratic differential

(¢ l Q(r)).

(4)

Fromfll(Az) = All fll(z)

with Il

(5)

t v we obtain after differentiation

i[v] 0 A = A[v] + A'f[v]


and hence Az

(the existence of A requires some, but not much, proof). Since

i [v lz = v differentiation of (5) gives

(v 0 A) A' = A- + A'vz

o because v ( B (i). We conclude that the. .A are analytic functions. Moreover, AIA' is real on the real

axis and can hence be extended by symmetry to the whole plane.

The explicit formula for i[v 1 shows that it is o(lz 12 ) at 00.

The apparent singularity of AIA' at A-loo is therefore re

movable, and there is at most a double pole at 00. We conclude

that

(6)

where PAis a second degree polynomial. From

(AlA

2)

(AlA

2)'

we deduce that

(AI 0 A2

) + (AI' 0 A2

) A2

(AI' 0 A2

) A2

'

(7)A'

2

We shall say that v is trivial, and we write v ( N (i) ,

if all A[v] = O. There is a whole slew of equivalent condi-

tions:

LEMMA 1. The following conditions are all equivalent.

a) A[v] = 0 for all A ( i, c) i[v] = 0 on R,

b) PA = 0 for all A ( i, d) cI> [v] == 0,

e) ¢[v1 == 0,

f) II v¢ dxdy = 0 for all ¢ ( Q(['). *i\H


Proof. a) < >b) by (6). c) > a) by (5). Con-

versely, if A '" 0 then since ;(0) = 0 it follows that ;(A 0)

= 0 for all A. .These points are dense on R (we are assuming

that the group is of the first kind) and hence, by continuity, f = 0

on R.

The definition of cI>, together with (5) and (6), gives

cI> ;, A - cI> = PA [v] + i PA [i v J

If cI> =0 0, it follows that PA [v] + i PA[i v] '" 0 on R. But

both polynomials are real on R, hence identically zero, so

d ) > b). Conversely, if f [v] = 0 on R, then cI> is pure-

ly imaginary on R and can be extended to be analytic in the

whole plane. Since cI> '" 0 (j z 12) it is a first degree polynomial.

But it vanishes at 0 and 1. Hence cI> =0 0 and c) > d).

Since cI> '" '" ¢, d) > e). Conversely, if ¢ = 0,

cI> is a polynomial and we reason as above to conclude cI> =0 o.Condition f) is the most important one. We prove it

only for the case that r\H is compact, and we represent it as

a compact fundamental polygon S with matched sides. If A =.. .0, then f 0 A = A'f. We know further that f z = v. By

Stokes' formula we find

f fv¢ dxdy

s- ii f f;z ¢ dz dz

s2~ f f ¢ dz

as

* For r\H non-compact, the condition shall hold for all ¢ f Q(r)

such that ffr\H 1¢ I dxdy < 00.


But the condition means that

f ¢ dz =

f/dz

fdz .

is invariant. Hence

is invariant, and the boundary integral vanishes.

To prove the opposite, consider a real ~ so that (3)

takes the form

~ -~ !!v (z ) R (z, ~) dx dy

H

We introduce a Poincare O-series

From the fact that IH IR (z, ~) Idx dy < 00 it is quite easy to

deduce that the series converges. We obtain now

~ = - ~ L: !f v(z)R(z, ~)dxdyA(S)

- ~ L: J!v(Az)R(Az, ~)IA '(z)1 2 dxdy

S

-~ L: Jf v(z)R(Az, ~)A'(z)2dxdy

= - ~ ff v !/J dx dy

which is zero if f) holds. Hence <I> is identically zero. The

lemma is proved.

We need the following lemma.

142

Then

(8)


LEMMA 2. Suppose ¢ is analytic in Hand

sup I¢ ly2 < 00 •

12 II p(z)y2 dxdy .17 (z _~)4

H

For the proof we note that

_ V. (z-z)2 = _ %[_1__ 2(z-C) +(z _~)4 ~z _~)2 (:z _~)3

_ 14~ [__1_ + z-( _ ~ (z-<:yld:Z (:z-~) (z_~)2 (z-~)3J

Assume first that ¢ is still analytic on R. Then integration

by parts gives

12 I I ¢ i dx dy = _ L17 (:z _ ~)4 217i

H RIt is easy to complete the proof by applying this formula to

¢(z+iE), E > 0, for the hypothesis guarantees absolute

convergence.

Compare formula (8) with (4). We have defined an anti

linear mapping

A: v ~ ¢[v1from B (r) to Q(r). On the other hand we may define

A*: ¢ ~ -¢ iwhich is a mapping from Q(r) to B (r). Lemma 2 tells us

*that AA is the identity.

By Lemma 1, e), v f N(r) if and only if Av = o.From AA* = I we conclude that

TeichmrlIIer Spaces 143

*v - A Av £ N (r) .

In other words, v is equivalent modN(r) to -¢;[vli. Of

*course this is the only A ¢ which is equivalent to v, for if

- ¢ y2 £ N (r) then fJhH 1¢ (z ) 12 y2 dx dy = 0, hence ¢ =

o.We conclude:

A establishes an isomorphism of B <r)/N (r) on Q(r) .

The inverse isomorphism of Q(i) on B(r)/N(r) is given by

A*.

A compact surface S, of genus g > Q, determines a

group i generated by linear transformations AI' ... , A2el

which satisfy

(9) ~ A2

Al

- l A2-1 ... A

2el-

lA

2elA

2el-

2-1 A

2el- l = I.

We say that lA l , ..• , A2el

l is a canonical set. If it belongs to

a surface, Al and A2 have four distinct fixpoints. By passing

to a conjugate subgroup we can make Al have fixpoints at 0,

00 and A2

to have a fixpoint at 1. When this is so we say that

the generating system is normalized.

Set

V set of normalized canonical systems,

T set of normalized canonical systems

that come from a surface of genus g.

It can be shown that V is a real analytic manifold of

dimension 6g - 6 . We are going to prove that T is an open sub

set of V, and that it carries a natural complex structure.


o all j

Let S determine r with normalized generators (A)

(A 1, ... , A2~)' We choose a basis v 1' ... , v3~-3 of

B (r)lN (r) and set

vCT) = T1V1 + .,. + T3~3 v3

and Tk = tk + itk'. For small T we obtain a system (A)V(T)

( T. The points of the manifold V near (A) can be expressed

by local parameters u1' ... , u6~6' and the mapping T ~ (A )V(T)

takes the form

uj = h j (tl' t 1 ', ... , t3~-3' t;g-3) .

We have to show

1) The h. are continuously differentiable,}

2) The Jacobian is f. 0 at T = O.

1) has already been proved. The coefficients of the A k

are differentiable functions of the uk' Therefore, if all Uk[v]

are 0, so are all A k [V] and hence all A[v]. Observe that

ah. ahj} = U. [Vk] - = u}.[ivk]

atk

} atk '

If the Jacobian at the origin were zero there would exist real

numbers (k' TJk such that

ah. ah j

~ (k at~ + TJk atk

'

This would mean

and hence

Hence ~ «(k + iTJk) v k ( N (r), and this is possible only if


A/1 .

all ~k' TJ k = o. We have proved that the Jacobian does not

vanish.

The proof shows that T is an open subset of V. It al

so follows that if 11/111 is small enough, then there exist unique

complex numbers T1(/1), , T3g-3(/1) such that

AT1 (/1)v1+ + T3/1-3 (/1)v3/1-3

Set /1 = tp and differentiate with respect to t for t = o.We obtain

A[~1[plv1 + ... + ~3~-3[p]v3g-31 A[p]

This implies

~1[plv1 + ... + ~3~-3[plv3~-3 - P £. N(r) .

Replace p by i p. We can then eliminate the term p to obtain

3~-3

L: (~k[P] + i ~k[ip])vk £. N(r)

1

from which it follows that

~k[i P1 = i ~k[P ]

In other words, the ~k are complex linear functionals,

which means that the Tk(/1) are differentiable in the complex

sense at /1 = o.From this we can prove that the coordinate mappings

(A/1) .... (T1(/1), ... , T3g-3(/1)) define a complex structure on T.

Indeed, we must show that on overlapping neighborhoods the

coordinates are analytic functions, and it suffices to show this

at the origin of a given coordinate system. Take /1 (T) =

l TjVj and /10 = /1(TO) near zero in B(r). Define A(r) in


B <r 11 0 ) by f 11 (r) = f'\ (t)

C in Chapter I, we have

11 .o f o. By the formulas of section

so that A depends analytically on T.

Now choose a basis A1

, ... , A3~-3 for B(r llo )/ N(r llo) . .;

Near (A 11 0) we have coordinate functions a 1(A), ... , a3~_3(A).

This means that for T near TO we can write uniquely

(AIl(r)) = (AIlO)A(r)) = (AIlO/a/A(r)) A)Because a/A) is complex analytic at A ". 0, we conclude )

that a i is a complex analytic function of T at TO' This is

exactly what we required.

.J

VAN NOSTRAND MATHEMATICAL STUDIES

VAN NOSTRAND MATHEMATICAL STUDIES are paperbacksfocusing on the living and growing aspects of mathematics. Theyare not reprints, but original publications. They are intended toprovide a setting for experimental, heuristic and informal writingin mathematics that may be research or may be exposition.

Under the editorship of Paul R. Halmos and Frederick W. Gehring,lecture notes, trial manuscripts, and other informal mathematicalstudies will be published in inexpensive, paperback format.

PAUL R. HALMOS graduated from the University of Illinois in1934. He took his master's degree in 1935 and his doctorate in 1938,and stayed on for another year to teach at the University. He thenspent three years at the Institute for Advanced Study, two of themas assistant to John von Neumann. He has returned to the Instituteon two subsequent occasions, in 1947-48 as a Guggenheim fellowand in 1957-58 on sabbatical leave. After teaching at SyracuseUniversity for three years he joined the faculty of the Universityof Chicago in 1946; he is now Professor of Mathematics at theUniversity of Michigan.

FREDERICK W. GEHRING attended The University of Michigan from 1943 to 1949, and received his doctorate from the University of Cambridge in 1952 after three years' study in England. Hespent the next three years as a Benjamin Pierce Instructor atHarvard University, and returned to The University of Michiganin 1955. From 1958-59 Professor Gehring was a Guggenheim andFulbright fellow at the University of Helsinki, Finland, and thefollowing year was an N.S.F. fellow at the Eidgenossische Technische Hochshule in Zurich, Switzerland. He is now Professor ofMathematics at the University of Michigan.

A VAN NOSTRAND MATHEMATICAL STUDY

under the general editorship of

PAUL R. HALMOS and FREDERICK W. GEHRING

The University of Michigan

•About this book:

In essentially self-contained exposition, this book presents the firsttreatment in English of a subject rapidly growing in its interest andimportance to mathematicians: German works in the same fieldoverlap only partly. Intended for advanced graduate study, thisvolume derives from lectures delivered by Professor Ahlfors atHarvard University during the Spring Term of 1964.

•The Author:

LARS V. AHLFORS is William Caspan Graustein Professor ofMathematics at Harvard University. As well as serving as chairmanof Harvard's department of mathematics from 1948 to 1950, theauthor has held professorships at the University of Helsingforsand the University of Zurich. A renowned mathematician, Professor Ahlfors was awarded the Field's medal for mathematicalresearch at the 1936 International Congress of Mathematicians; heis a member of numerous distinguished societies, including theNational Academy of Science. Professor Ahlfors has contributedmuch to the development of conformal mapping,. Riemann surfaces, and other branches of the theory of functions of a complexvariable: he is the author of Complex Analysis and, with L. Sario, ofRiemann Surfaces.

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ahlfors l.v. - lectures on quasi con formal mappings (van nostrand, 1966)

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